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Concentration Definition By: Anne Marie Helmenstine, Ph.D. Updated on February 12, 2020 In chemistry, the word "concentration" relates to the components of a mixture or solution. Here is the definition of concentration and a look at different methods used to calculate it. Concentration Definition In chemistry, concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution. Concentration is usually expressed in terms of mass per unit volume. However, the solute concentration may also be expressed in moles or units of volume. Instead of volume, concentration may be per unit mass. While usually applied to chemical solutions, concentration may be calculated for any mixture. Unit Examples of Concentration: g/cm3, kg/l, M, m, N, kg/L How to Calculate Concentration Concentration is determined mathematically by taking the mass, moles, or volume of solute and dividing it by the mass, moles, or volume of solution (or, less commonly, the solvent). Some examples of concentration units and formulas include: • • • • • • • • • • • • • • Molarity (M) - moles of solute/litres of solution (not solvent!) Mass Concentration (kg/m3 or g/L) - mass of solute/volume of solution Normality (N) - grams active solute/litres of solution Molality (m) - moles of solute/mass of solvent (not mass of solution!) Mass Percent (%) - mass solute/mass solution x 100% (mass units are the same unit for both solute and solution) Volume Concentration (no unit) - volume of solute/volume of mixture (same units of volume for each) Number Concentration (1/m3) - number of entities (atoms, molecules, etc.) of a component divided by the total volume of the mixture Volume Percent (v/v%) - volume solute/volume solution x 100% (solute and solution volumes are in the same units) Mole Fraction (mol/mol) - moles of solute/total moles of species in the mixture Mole Ratio (mol/mol) - moles of solute/total moles of all other species in the mixture Mass Fraction (kg/kg or parts per) - mass of one fraction (could be multiple solutes)/total mass of the mixture Mass Ratio (kg/kg or parts per) - mass of solute/mass of all other constituents in the mixture PPM (parts per million) - a 100 ppm solution is 0.01%. The "parts per" notation, while still in use, has largely been replaced by mole fraction PPB (parts per billion) - typically used to express contamination of dilute solutions Some units may be converted from one to another. However, it's not always a good idea to convert between units based on the volume of solution to those based on mass of solution (or vice versa) because volume is affected by temperature. Strict Definition of Concentration In the strictest sense, not all means of expressing the composition of a solution or mixture fall under the simple term "concentration". Some sources only consider mass concentration, molar concentration, number concentration, and volume concentration to be true units of concentration. Concentration Versus Dilution Two related terms are concentrated and dilute. Concentrated refers to chemical solutions that have high concentrations of a large amount of solute in the solution. If a solution is concentrated to the 1 point where no more solute will dissolve in the solvent, it is said to be saturated. Dilute solutions contain a small amount of solute compared with the amount of solvent. In order to concentrate a solution, either more solute particles must be added or some solvent must be removed. If the solvent is non-volatile, a solution may be concentrated by evaporating or boiling off solvent. Dilutions are made by adding solvent to a more concentrated solution. It's common practice to prepare a relatively concentrated solution, called a stock solution, and use it to prepare more dilute solutions. This practice results in better precision than simply mixing up a dilute solution because it can be difficult to obtain an accurate measurement of a tiny amount of solute. Serial dilutions are used to prepare extremely dilute solutions. To prepare a dilution, stock solution is added to a volumetric flask and then diluted with solvent to the mark. Source: IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). 2 Concentration Definition By: Anne Marie Helmenstine, Ph.D. Updated on February 12, 2020 In chemistry, the word "concentration" relates to the components of a mixture or solution. Here is the definition of concentration and a look at different methods used to calculate it. Concentration Definition In chemistry, concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution. Concentration is usually expressed in terms of mass per unit volume. However, the solute concentration may also be expressed in moles or units of volume. Instead of volume, concentration may be per unit mass. While usually applied to chemical solutions, concentration may be calculated for any mixture. Unit Examples of Concentration: g/cm3, kg/l, M, m, N, kg/L How to Calculate Concentration Concentration is determined mathematically by taking the mass, moles, or volume of solute and dividing it by the mass, moles, or volume of solution (or, less commonly, the solvent). Some examples of concentration units and formulas include: • • • • • • • • • • • • • • Molarity (M) - moles of solute/litres of solution (not solvent!) Mass Concentration (kg/m3 or g/L) - mass of solute/volume of solution Normality (N) - grams active solute/litres of solution Molality (m) - moles of solute/mass of solvent (not mass of solution!) Mass Percent (%) - mass solute/mass solution x 100% (mass units are the same unit for both solute and solution) Volume Concentration (no unit) - volume of solute/volume of mixture (same units of volume for each) Number Concentration (1/m3) - number of entities (atoms, molecules, etc.) of a component divided by the total volume of the mixture Volume Percent (v/v%) - volume solute/volume solution x 100% (solute and solution volumes are in the same units) Mole Fraction (mol/mol) - moles of solute/total moles of species in the mixture Mole Ratio (mol/mol) - moles of solute/total moles of all other species in the mixture Mass Fraction (kg/kg or parts per) - mass of one fraction (could be multiple solutes)/total mass of the mixture Mass Ratio (kg/kg or parts per) - mass of solute/mass of all other constituents in the mixture PPM (parts per million) - a 100 ppm solution is 0.01%. The "parts per" notation, while still in use, has largely been replaced by mole fraction PPB (parts per billion) - typically used to express contamination of dilute solutions Some units may be converted from one to another. However, it's not always a good idea to convert between units based on the volume of solution to those based on mass of solution (or vice versa) because volume is affected by temperature. Strict Definition of Concentration In the strictest sense, not all means of expressing the composition of a solution or mixture fall under the simple term "concentration". Some sources only consider mass concentration, molar concentration, number concentration, and volume concentration to be true units of concentration. Concentration Versus Dilution Two related terms are concentrated and dilute. Concentrated refers to chemical solutions that have high concentrations of a large amount of solute in the solution. If a solution is concentrated to the 1 point where no more solute will dissolve in the solvent, it is said to be saturated. Dilute solutions contain a small amount of solute compared with the amount of solvent. In order to concentrate a solution, either more solute particles must be added or some solvent must be removed. If the solvent is non-volatile, a solution may be concentrated by evaporating or boiling off solvent. Dilutions are made by adding solvent to a more concentrated solution. It's common practice to prepare a relatively concentrated solution, called a stock solution, and use it to prepare more dilute solutions. This practice results in better precision than simply mixing up a dilute solution because it can be difficult to obtain an accurate measurement of a tiny amount of solute. Serial dilutions are used to prepare extremely dilute solutions. To prepare a dilution, stock solution is added to a volumetric flask and then diluted with solvent to the mark. Source: IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). 2 How do you calculate molarity? Molarity is calculated, using the molarity formula above, by considering two components: volume and moles. In the case that moles of the compound are unknown, molar mass can be used to convert the compound from grams to moles. The periodic table provides the atomic masses that are used to calculate molar mass. Step 1: The first step to calculating molarity is identifying one of the two key factors that make up the solution: the volume of the solution and the amount of solute in grams or moles. First, we will start with volume in this tutorial. The volume of the solution can be measured by using a graduated cylinder. For molarity, volume must be in the unit of liters. If the starting volume is in milliliters, it must be converted to liters before calculating molarity. There are 1000 millimeters in a liter. So, with a simple calculation, any volume in milliliters can be converted to liters. For example, if the volume of the solution is 100 mL: Step 2: The second step is to determine the amount of solute present in the solution in moles. If the known amount of solute is in grams, it must be converted to moles using molar mass. If we say that the solute is 5.00 g of ammonia (NH3), we can convert this to moles using ammonia’s molar mass (17.04 g/mol): Step 3: The third and final step is to use the molarity formula and divide the number of moles of solute by the number liters of the solution to obtain the molarity in moles per liter. If we take the two values from the previous step, we see that the ammonia solution is 2.9 M. This means that every liter of this solution contains 2.9 moles of ammonia. Another example of calculating molarity Using a different compound, calcium chloride, we can calculate the molarity of a solution in the same way. Let’s start with these values: • 10.0 g CaCl2 • 200 mL H2O Following the same process outlined above, we can determine the molarity of this calcium chloride solution in a few simple steps. First, the volume must be converted from milliliters to liters. 1 Next, we convert grams of calcium chloride into moles. Finally, we divide the number of moles by the volume of the solution. How do you calculate osmolarity from molarity? You multiply the molarity by the number of osmoles that each solute produces. Explanation: An osmole (Osmol) is 1 mol of particles that contribute to the osmotic pressure of a solution. For example, NaCl dissociates completely in water to form Na+ ions and Cl- ions. Thus, each mole of NaCl becomes two osmoles in solution: one mole of Na+ and one mole of Cl-. A solution of 1 mol/L NaCl has an osmolarity of 2 Osmol/L. Also, a solution of 1 mol/L Calcium dichloride (CaCl2) has an osmolarity of 3 Osmol/L (1 mol Ca2+ and 2 mol Cl-). EXAMPLE Calculate the osmolarity of blood. Solution The concentrations of solutes are: [Na+] = 0.140 mol/L; [glucose] = 180 mg/100 mL; [BUN] (blood urea nitrogen) = 20 mg/100 mL. [Na+]= 0.140 mol/L. But, each Na+ ion pairs with a negative ion X- such as Cl- to give 2 Osmol of particles. NaX osmolarity = 2 Glucose osmolarity= BUN osmolarity= Blood osmolarity = How can molarity and osmolarity be calculated from mass per unit volume? Molarity is the number of moles of solute per litre of solution; osmolarity is the number of osmoles per litre of solution. Assume you have a solution that contains 40.0 g of NaCl in 1.00 L of solution. MOLARITY The molarity of a solution is the number of moles of solute per litre of solution. M = n/V You have to convert the mass of NaCl to moles of NaCl. OSMOLARITY An ionic compound such as NaCl separates into cations and anions on dissolving. NaCl(s) →Na⁺(aq) + Cl⁻(aq) Osmosis depends on the number of solute particles in a solution. Both the number of moles of Na+ and the number of moles of Cl- contribute to the osmotic pressure. Thus, 1 mol of NaCl gives 2 mol of osmotic particles. 3 Each mole of solute that contributes to the osmotic pressure of a solution is an osmole. The osmolarity is the molarity of the compound times the number of particles produced by one mole of compound. The osmolarity of the NaCl solution is 4 How do you calculate molarity? Molarity is calculated, using the molarity formula above, by considering two components: volume and moles. In the case that moles of the compound are unknown, molar mass can be used to convert the compound from grams to moles. The periodic table provides the atomic masses that are used to calculate molar mass. Step 1: The first step to calculating molarity is identifying one of the two key factors that make up the solution: the volume of the solution and the amount of solute in grams or moles. First, we will start with volume in this tutorial. The volume of the solution can be measured by using a graduated cylinder. For molarity, volume must be in the unit of liters. If the starting volume is in milliliters, it must be converted to liters before calculating molarity. There are 1000 millimeters in a liter. So, with a simple calculation, any volume in milliliters can be converted to liters. For example, if the volume of the solution is 100 mL: Step 2: The second step is to determine the amount of solute present in the solution in moles. If the known amount of solute is in grams, it must be converted to moles using molar mass. If we say that the solute is 5.00 g of ammonia (NH3), we can convert this to moles using ammonia’s molar mass (17.04 g/mol): Step 3: The third and final step is to use the molarity formula and divide the number of moles of solute by the number liters of the solution to obtain the molarity in moles per liter. If we take the two values from the previous step, we see that the ammonia solution is 2.9 M. This means that every liter of this solution contains 2.9 moles of ammonia. Another example of calculating molarity Using a different compound, calcium chloride, we can calculate the molarity of a solution in the same way. Let’s start with these values: • 10.0 g CaCl2 • 200 mL H2O Following the same process outlined above, we can determine the molarity of this calcium chloride solution in a few simple steps. First, the volume must be converted from milliliters to liters. 1 Next, we convert grams of calcium chloride into moles. Finally, we divide the number of moles by the volume of the solution. How do you calculate osmolarity from molarity? You multiply the molarity by the number of osmoles that each solute produces. Explanation: An osmole (Osmol) is 1 mol of particles that contribute to the osmotic pressure of a solution. For example, NaCl dissociates completely in water to form Na+ ions and Cl- ions. Thus, each mole of NaCl becomes two osmoles in solution: one mole of Na+ and one mole of Cl-. A solution of 1 mol/L NaCl has an osmolarity of 2 Osmol/L. Also, a solution of 1 mol/L Calcium dichloride (CaCl2) has an osmolarity of 3 Osmol/L (1 mol Ca2+ and 2 mol Cl-). EXAMPLE Calculate the osmolarity of blood. Solution The concentrations of solutes are: [Na+] = 0.140 mol/L; [glucose] = 180 mg/100 mL; [BUN] (blood urea nitrogen) = 20 mg/100 mL. [Na+]= 0.140 mol/L. But, each Na+ ion pairs with a negative ion X- such as Cl- to give 2 Osmol of particles. NaX osmolarity = 2 Glucose osmolarity= BUN osmolarity= Blood osmolarity = How can molarity and osmolarity be calculated from mass per unit volume? Molarity is the number of moles of solute per litre of solution; osmolarity is the number of osmoles per litre of solution. Assume you have a solution that contains 40.0 g of NaCl in 1.00 L of solution. MOLARITY The molarity of a solution is the number of moles of solute per litre of solution. M = n/V You have to convert the mass of NaCl to moles of NaCl. OSMOLARITY An ionic compound such as NaCl separates into cations and anions on dissolving. NaCl(s) →Na⁺(aq) + Cl⁻(aq) Osmosis depends on the number of solute particles in a solution. Both the number of moles of Na+ and the number of moles of Cl- contribute to the osmotic pressure. Thus, 1 mol of NaCl gives 2 mol of osmotic particles. 3 Each mole of solute that contributes to the osmotic pressure of a solution is an osmole. The osmolarity is the molarity of the compound times the number of particles produced by one mole of compound. The osmolarity of the NaCl solution is 4 Examples of Osmolarity calculations. In some cases, the calculated osmolarity of a commercially available preparation is greater or less than what is printed on the vial. In those cases, the manufacturer’s data is used in the Osmolarity Determination Program. Calcium chloride: Exists as dihydrate (CaCl2.2h20). Total MW= (40.1 + 35.5 + 35.5+ (2×2) + (2×16)= 146.1 — Calculations: (1gram CaCl2/146.1) x 1000 x valence (2)= 13.68 meq (vial states: 13.6). Osmolarity: 1 gram/10ml or (100g/liter)/146.1 x 1000 = 684.46 mmol x #species (3) = 2053 mOsm/L (vial states: 2040 mOsm/L) Calcium gluconate: Osmolarity: 680 mOsm/L. MW: 430.38 concentration: 1 gram= 4.65 meq 9.3mg of calcium/ml. Calculation: 1 g/10ml or 100g/L — (100/430.38) x 1000 x # species (3 — ca(gluc)2 = 697 mOsm/L (calculated) Common IV D10: 505 mOsmol/L; NS: 308 mOsmol/L; LR: 273 mOsmol/L; D5LR: 525 solutions mOsmol/L; D5.45: 406 mOsmol/L; D5NS: 560 mOsmol/L; 0.45NS: 154 mOsmol/L Dextrose (glucose) Exists as a monohydrate: MW=180 + 18=198. 5% dext= 5g/100ml or 50g/1000ml. 50g/198=0.252 moles or 252 mmol x species (=1), therefore osmolarity=252 mOsm/L. D70: 700g/1000ml 700/198 x 1000=3535 mOsm/L (3530 bag). Magnesium sulfate: Exists as magnesium sulfate heptahydrate – MgSO4.7H2O MW=246.47 Calculations: (1gram MgSO4/246.47) x 1000= 4.057 or ~ 4.06 mmol x valence (2) = 8.12 meq. Osmolarity: 0.5g/ml or 500g/liter, therefore (500/246.47) x 1000=2028 mmol x # species (2) = 4057 mOsm/liter Potassium acetate: CH3COOK MW=98.1 Calculation: 2 meq/ml [((0.196 g/ml) x 1000)/98.1] x 1000= 2000 mMol x # of species (2) = 4000 mOsm/L. Potassium chloride: 149 mg/ml. MW=(39.1 + 35.5 = 74.6). 149/74.6 = 2 mmol x valence (1) = 2 meq/ml. Osmolarity: ((149g/L) / 74.6) x 1000 x # species(2) = ~ 4000 mOsm/L Potassium phosphate: Each ml contains 224mg of monobasic kphos (KH2PO4), and 236mg of dibasic kphos (K2HPO4). 93 mg phosphorus/ml = 3 mMol. 170.3 mg K+/ml = 4.4 meq. Osmolarity: 7400 mOsm/L. Sodium acetate: CH3cooNa MW=82. Calculations: 2 meq/ml solution (164 mg/ml) (0.164g x 1000)/82 x 1000= 2000 mMol x 2 species= 4000 mOsm/L. Sodium (NaHCO3 – MW=84). Calculations: 8.4% soln = 8.4 g/100 ml (8.4/84g) x 1000 bicarbonate: = 100 mmol x valence (1) = 100 meq or 1 meq/ml. Osmolarity: (84g/84g) x 1000 x # species (2) = 2000 mOsm/L. 7.5 % soln (~1790 mosm/L) Sodium chloride: Osmolarity: 8000 mOsm/L. MW: 58.5 Concentration: 23.4% solution. Calculations: 23.4g/100ml or 234 mg/ml. 234/58.5 = 4.0 x valence (1) = 4 meq/ml. Osmolarity: 234g/Liter = 4 moles or 4000 mmol x species (2) = 8000 mOsm/liter Sodium phosphate: Each ml contains 276mg of monobasic Naphos (NaH2PO4), and 142mg of dibasic naphos (Na2HPO4). 93 mg of phosphorus/ml = 3 mmol. 92 mg of sodium per ml/23 = 4 meq/ml. Osmolarity: 7000 mOsm/L TPN Solutions Aminosyn II 15% amino acid: 1300 mOsm/L; Travasol 10% amino acid: 998 mOsm/L; Lipid 10%: 276 mOsm/Liter 20%: 258 mOsm/L. Novamine 15%: 1388 mOsm/L; HepatAmine 8% amino acid: 785 mOsmol/L. Normality Normality of solution= equivalents per liter or meq per ml. // # of moles of solute x valence= normality. Therefore a 1 molar solution of HCl= 1 normality. or a 1 molar CaCl2 solution = 2N cacl2 Examples of Osmolarity calculations. In some cases, the calculated osmolarity of a commercially available preparation is greater or less than what is printed on the vial. In those cases, the manufacturer’s data is used in the Osmolarity Determination Program. Calcium chloride: Exists as dihydrate (CaCl2.2h20). Total MW= (40.1 + 35.5 + 35.5+ (2×2) + (2×16)= 146.1 — Calculations: (1gram CaCl2/146.1) x 1000 x valence (2)= 13.68 meq (vial states: 13.6). Osmolarity: 1 gram/10ml or (100g/liter)/146.1 x 1000 = 684.46 mmol x #species (3) = 2053 mOsm/L (vial states: 2040 mOsm/L) Calcium gluconate: Osmolarity: 680 mOsm/L. MW: 430.38 concentration: 1 gram= 4.65 meq 9.3mg of calcium/ml. Calculation: 1 g/10ml or 100g/L — (100/430.38) x 1000 x # species (3 — ca(gluc)2 = 697 mOsm/L (calculated) Common IV D10: 505 mOsmol/L; NS: 308 mOsmol/L; LR: 273 mOsmol/L; D5LR: 525 solutions mOsmol/L; D5.45: 406 mOsmol/L; D5NS: 560 mOsmol/L; 0.45NS: 154 mOsmol/L Dextrose (glucose) Exists as a monohydrate: MW=180 + 18=198. 5% dext= 5g/100ml or 50g/1000ml. 50g/198=0.252 moles or 252 mmol x species (=1), therefore osmolarity=252 mOsm/L. D70: 700g/1000ml 700/198 x 1000=3535 mOsm/L (3530 bag). Magnesium sulfate: Exists as magnesium sulfate heptahydrate – MgSO4.7H2O MW=246.47 Calculations: (1gram MgSO4/246.47) x 1000= 4.057 or ~ 4.06 mmol x valence (2) = 8.12 meq. Osmolarity: 0.5g/ml or 500g/liter, therefore (500/246.47) x 1000=2028 mmol x # species (2) = 4057 mOsm/liter Potassium acetate: CH3COOK MW=98.1 Calculation: 2 meq/ml [((0.196 g/ml) x 1000)/98.1] x 1000= 2000 mMol x # of species (2) = 4000 mOsm/L. Potassium chloride: 149 mg/ml. MW=(39.1 + 35.5 = 74.6). 149/74.6 = 2 mmol x valence (1) = 2 meq/ml. Osmolarity: ((149g/L) / 74.6) x 1000 x # species(2) = ~ 4000 mOsm/L Potassium phosphate: Each ml contains 224mg of monobasic kphos (KH2PO4), and 236mg of dibasic kphos (K2HPO4). 93 mg phosphorus/ml = 3 mMol. 170.3 mg K+/ml = 4.4 meq. Osmolarity: 7400 mOsm/L. Sodium acetate: CH3cooNa MW=82. Calculations: 2 meq/ml solution (164 mg/ml) (0.164g x 1000)/82 x 1000= 2000 mMol x 2 species= 4000 mOsm/L. Sodium (NaHCO3 – MW=84). Calculations: 8.4% soln = 8.4 g/100 ml (8.4/84g) x 1000 bicarbonate: = 100 mmol x valence (1) = 100 meq or 1 meq/ml. Osmolarity: (84g/84g) x 1000 x # species (2) = 2000 mOsm/L. 7.5 % soln (~1790 mosm/L) Sodium chloride: Osmolarity: 8000 mOsm/L. MW: 58.5 Concentration: 23.4% solution. Calculations: 23.4g/100ml or 234 mg/ml. 234/58.5 = 4.0 x valence (1) = 4 meq/ml. Osmolarity: 234g/Liter = 4 moles or 4000 mmol x species (2) = 8000 mOsm/liter Sodium phosphate: Each ml contains 276mg of monobasic Naphos (NaH2PO4), and 142mg of dibasic naphos (Na2HPO4). 93 mg of phosphorus/ml = 3 mmol. 92 mg of sodium per ml/23 = 4 meq/ml. Osmolarity: 7000 mOsm/L TPN Solutions Aminosyn II 15% amino acid: 1300 mOsm/L; Travasol 10% amino acid: 998 mOsm/L; Lipid 10%: 276 mOsm/Liter 20%: 258 mOsm/L. Novamine 15%: 1388 mOsm/L; HepatAmine 8% amino acid: 785 mOsmol/L. Normality Normality of solution= equivalents per liter or meq per ml. // # of moles of solute x valence= normality. Therefore a 1 molar solution of HCl= 1 normality. or a 1 molar CaCl2 solution = 2N cacl2 Osmolarity and Molarity The molarity of a solution is equal to the number of moles of the specified substance that are dissolved in one liter of solution (1000 ml of solution). Therefore, we have in 1000 ml solution ...200 mM in 300 ml solution......x mM x =300*200/1000 =60 mM How Can I Prepare 3 Molar KCL Solution For Volume Of 500 Ml? The molar mass of KCl is 74.5 (39+35.5). This means that a 1molar solution would contain 74.5g KCl per liter. Therefore, a 3 molar solution would contain 74.5 x 3 = 223.5g KCl per liter. Now, 1 liter has 1000 ml in it. We need to prepare a 3M solution in 500 ml. Just simple math here. If 1000ml has 223.5g of KCl in it to make a 3M solution, 500ml will have half the amount (since 500 is the half of 1000). So, we'll divide 223.5 by 2. This gives us 111.75. So, first of all you need to weigh out 111.75g of KCl. You do this by first weighing the empty dish on the scale and calibrating the scale to zero with the dish on it. Then add KCl to the dish on the scale until the scale reads 111.75g. How would you prepare 300 mL of 0.750 M KCl? 16.7 grams of KCl diluted to a final volume of 300mL. Explanation: Molarity is equal to the number of moles of a solute per 1 liter of solution. (It's a way to measure concentration in chemistry). In this case, the KCl is the solute, and 0.750 M tells you how many moles of it we have per 1 liter of solution. Convert moles of KCl to grams using its molar mass, 74.5513 g/mol. 0.750mol x (74.5513g/1mol) = 55.883 grams. 1 In one liter of solution, there are 55.883 grams of KCl. But the question wants to know about 300mL (which is equal to 0.3L) 0.3L x 55.883g/1L = 16.7499 grams, round to 3 SF = 16.7 grams. If a solution is 0.5% (w/v), how many mg/ml is that? 5 mg mL−1 The solution's mass by volume percent concentration, m/v %, tells you the number of grams of solute present for every 100 mL of the solution. In your case, the solution is said to have a mass by volume percent concentration of 0.5%, which means that you get 0.5 g of solute for every 100 mL of the solution. 0.5% m/v=0.5 g solute per 100 mL of the solution You can thus say that 1 mL of this solution will contain 1mL solution *0.5 g solute/100mL solution =0.005 g solute To convert this to milligrams, use the fact that 1 g=103 mg You will end up with 0.005g*103 mg/1g = 5 mg Since this represents the number of milligrams of solute present for every 1 mL of the solution, you can say that the solution has a concentration of concentration = 5 mg mL−1. The answer is rounded to one significant figure. 2 Convert moles of a substance into grams A mole is the quantity of anything that has the same number of particles found in 12.000 grams of carbon-12. That number of particles is Avogadro's Number, which is roughly 6.02x1023. One mole of atoms contains 6 x 1023 atoms, no matter what element it is. In order to convert moles of a substance into grams you need a periodic table or another list of atomic masses. How many grams of carbon dioxide is 0.2 moles of CO2? Look up the atomic masses of carbon and oxygen. This is the number of grams per one mole of atoms. Carbon (C) has 12.01 grams per mole. Oxygen (O) has 16.00 grams per mole. One molecule of carbon dioxide contains 1 carbon atom and 2 oxygen atoms, so: number of grams per mole CO2 = 12.01 + [2 x 16.00] = 12.01 + 32.00 = 44.01 gram/mole Now, multiply this number of grams per mole (44.01 gram/mole) multiply by the number of moles (0.2 moles) in order to get the final answer: grams in 0.2 moles of CO2 = 0.2 moles x 44.01 grams/mole = 8.802 grams How to Convert Milligrams Per Liter to Molarity 1 Calculate the molecular weight of the solute in the solution. For example, assume the solute is sodium chloride, which has the molecular formula NaCl. To find the molecular weight of sodium chloride, add the weight of Na (22.99) to Cl (35.45), which equals 58.44 grams per mole. 2 To convert milligrams per liter to molarity, use this formula: Molarity = mg/L / (molecular weight of solute) x 1,000. Example: the solute in the solution is sodium chloride and the concentration is 0.5 mg/L, the equation would be Molarity = 0.5 mg/L / (58.44) x 1,000. The answer is 8.556 micromoles (μM). Convert from micromoles to moles or millimoles if necessary. 3 To calculate moles, use the equation Moles = micromoles / 1,000,000. To calculate millimoles, use the equation Millimoles = micromoles / 1,000 How to Interconvert Moles, Molarity and Volume 1 Calculate the molarity of a solution in moles per liter, given moles and volume in liters, by dividing the number of moles by the volume. For example, a 5.0 liter solution containing 10.0 moles has a molarity of 2.0 moles per liter. 2 Determine the number of moles in a solution, with molarity and volume known, by multiplying molarity in moles per liter by the volume in liters Example: a 2.0 liter solution with a molarity of 3.0 moles per liter. There are 6.0 moles in the solution. 3 Compute the volume of a solution in liters, given the number of moles and molarity, by dividing the number of moles by the molarity in units of moles per liter. Example: a solution containing 6.0 moles and a having a molarity of 3.0 moles per liter has a volume of 2.0 moles per liter. Exercise Calculate the number of grams of calcium hydroxide in 1.30 L of a 0.75 M Ca(OH)2 solution. Moles of Ca(OH)2 = 0.75 M x 1.30 L = 0.975 mol/L of Ca(OH)2 How many moles of potassium chloride (KCl) are contained in 0.25 L of 0.80 mol/L KCl? Moles of KCl = molarity x volume = 0.80 mol/L KCl x 0.25 L =0.20 mol/L Exercise A 4 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 ml teacup filled with hot water. What is the molarity of the sugar solution? M = m/V where M is molarity (mol/L) m = number of moles of solute V = volume of solvent (Liters) Determine number of moles of sucrose in 4 g C12H22O11 = (12)(12) + (1)(22) + (16)(11) 4 C12H22O11 = 144 + 22+ 176 C12H22O11 = 342 g/mol Divide this amount into the size of the sample 4 g/(342 g/mol) = 0.0117 mol Determine the volume of solvent in liters 350 ml x (1L/1000 ml) = 0.350 L Determine the molarity of the solution M = m/V M = 0.0117 mol /0.350 L M = 0.033 mol/L Answer: The molarity of the sugar solution is 0.033 mol/L. Exercise Calculate the Osmolarity of 0.3% Saline Each 100 mL contains 300 mg sodium chloride in water. This is 3 g per L. The solution is 3 g of NaCl dissolved in water to a total volume of 1000 mL. The molecular weight of NaCL is 58.5g per mole  58.5g NaCl = 1 mole 0.3% NaCl contains 3 g NaCl/1L solution The concentration is 3g/L divided by 58.5g/mole = 0.051 mole per liter.  Molarity of 3g of NaCI = 3g/L / 58.5g/mol = 0.051 mol/L OR 51 mmol/L solution of NaCl Since each molecule of NaCl dissociates into Na+ and Cl− ions Osmolarity of 0.3% saline  Molar value (51) is multiplied by 2 = 51*2 = 102 mOsm/L. Exercise Calculate the Osmolarity of 0.9% Saline Each 100 mL contains 900 mg sodium chloride in water. This is 9 g per L. The solution is 9 g of NaCl dissolved in water to a total volume of 1000 mL. The molecular weight of NaCL is 58.5g per mole  58.5g NaCl = 1 mole 0.9% NaCl contains 9 g NaCl/1L solution The concentration is 9g/L divided by 58.5g/mole = 0.154 mole per liter. 5  Molarity of 9g of NaCI = 9g/L / 58.43g/mol = 0.154 mol/L OR 154 mmol/L solution of NaCl Since each molecule of NaCl dissociates into Na+ and Cl− ions Osmolarity of 0.9% saline  Molar value (154) is multiplied by 2 = 154*2 = 308 mOsm/L. Exercise Find the osmolarity of 150 mM Sodium chloride (NaCl) 150 mM Sodium chloride (NaCl) has an osmolarity of  150 mM Na+ + 150 mM Cl− = 300 mOsmol.  150 mM NaCl has an osmolarity of 300 mOsmol. Find the osmolarity of 50 mM Calcium chloride (CaCl2) 50 mM Calcium chloride (CaCl2) and 5 mM Sodium bicarbonate (NaHCO3) have an osmolarity of  (1 * 50 mM Ca2+ + 2 * 50 mM Cl−) + (1 * 5 mM Na+ + 1 * 5 mM HCO3−) = 160 mOsmol.  50 mM CaCl2 and 5 mM NaHCO3 have an osmolarity of 160 mOsmol. Exercise If a red blood cell (typically 300 mOsmol/L) was placed into a solution of 100 mmol/L of NaCl (so 200 mOsmol/L because NaCL dissociates into 2 ions in a solution) then the cell would swell (the extracellular solution is hypotonic), If it was placed into a solution of 200 mmol/L of NaCl (400 mOsmol/L) then the cell would shrink (the solution is hypertonic), and if it was placed into a solution of 150 mmol/L of NaCl (300 mOsmol/L) the size of the cell would not change (the extracellular solution is isotonic). Exercise The osmolarity of a solution is the total number of solutes within the solution and if it is hypo/hyper/isosmotic it is in comparison to the osmolarity of another solution. Therefore, 100 mmol/L of NaCl would have an osmolarity of 200mOsmol/L. 6 100 mmol/L of NaCl plus 100 mmol/L of urea would have an osmolarity of 300 mOsmol/L because you add up ALL of the particles. NaCl makes up 2 ions in water  2 Osm/L  100mmol/L NaCl = 200mOsmol/L Urea does not dissociate in water  1 Osm/L  100mmol/L Urea = 100mOsmol/L The total osmolarity of the solution is NaCl 300 + Urea 100 = 300 mOsmol Exercise In order to determine the state of hyper/hypo/isosmolarity hyper/hypo/isotonicity of a solution you need to be comparing it TO something. OR Typically, we compare a solution to the state of a cell, which has 300mOsmol/L osmolarity Note that cells are made up of particles and proteins that CANNOT move across the plasma membrane. Therefore, to determine the tonicity (hyper / hypo / isotonicity) of a solution compared to the inside of a cell you would look at for the number of particles in that solution that cannot cross the plasma membrane (ex. ions) and compare that number to 300 mOsmol/L. So, 100 mmol/L of NaCl (200 mOsmol/L) would be hypotonic. It is also hypo-osmotic because it is only 200mOsm compared to 300mOsm of the cell The osmolarity of a solution containing both NaCl and Urea would therefore be 400 mOsmol/L. If we were to compare the osmolarity of this solution to that of a typical cell (300 mOsmol/L), then it would be hyperosmotic (the solution has an osmolarity of 400 mOsmol/L while the cell has an osmolarity of 300. If we were to compare the tonicity of this solution to that of a typical cell (300 mOsmol/L), then it would be isotonic (both have tonicities of 300 mOsmol/L). Therefore our solution is hyperosmotic and isotonic 7 Exercise We have two solutions (A and B) separated by a membrane that is permeable to water but not to solutes. Solution A was prepared by dissolving 5g of sodium chloride in 10L of water. Solution B was prepared by dissolving 12g of sodium chloride in 6L of water. a) What is the osmolarity of side A in the initial condition before osmotic equilibrium has been established? b) What is the osmolarity of side B in the initial condition before osmotic equilibrium has been established? c) Which direction will water flow to establish osmotic equilibrium in the condition above? (From A to B, from B to A, or there will be no net movement of water?) A. Molecular weight of sodium chloride = 58.44g/mole 1.0 mole NaCl = 2.0 osmoles (5g NaCl/10L) x (mol/58.44g) X (2.0 osmole/mole) = 0.017 osmoles B. Molecular weight of sodium chloride = 58.44g/mole 1.0 mole NaCl = 2.0 osmoles (12g NaCl/6L) x (mol/58.44g) X (2.0 osmole/mole) = 0.068 osmoles C. Water follows the solute, so from A to B Exercise Two solutions (A and B) are separated by a membrane that is permeable to water but not to solutes. Solution A contains 400osmoles of sodium chloride dispersed in a 5L solution. Solution B contains 240osmoles of sodium chloride dispersed in a 12L solution. a) What is the osmolarity of side A in the initial condition before osmotic equilibrium has been established? b) What is the osmolarity of side B in the initial condition before osmotic equilibrium has been established? c) Which direction will water flow to establish osmotic equilibrium in the condition above? (From A to B, from B to A, or there will be no net movement of water?) A. 400 osmoles/5L = 80 osmoles/L = 80 OsM B. 240 osmoles/12L = 20 osmoles/L = 20 OsM C. From B to A 8 Exercise A 5 liter solution with 20 osmoles of sucrose (side A) is separated from a 2 liter solution with 10 osmoles of sucrose (side B) by a membrane that is permeable to water but impermeable to sucrose. a) Which direction will water flow? (From A to B, from B to A, or there will be no net movement of water?) Side A: 20 osmoles/5 L = 4 OsM. Side B: 10 osmoles/2L = 5 OsM b) Which direction will water flow if the membrane is permeable to both sucrose in water? (From A to B, from B to A, or there will be no net movement of water?) A. From side A to B B. The solute will move to equilibrium, which means that both sides will have 4.5 OsM. Therefore, there will be no net movement of water. Exercise A 300 mOsM urea solution is ………… hyposmotic to a 200 mM solution of NaCl. (200mM NaCl = 400mOsM NaCl) Tonicity is …………… A relative term that describes the effect a solution has on cells of the body. True or False: A hypertonic solution will cause a cell to swell. False True or False: A solution can be hyperosmotic and still be hypo- or isotonic True Exercise Red blood cells are placed in a beaker filled with 0.5L of a 150mM NaCl solution. NaCl is a non-penetrating solute and cannot cross the red blood cell membrane. a) What will happen to the red blood cells? b) Is the solution above hypotonic, hypertonic, or isotonic? A. What will happen to the red blood cells? Osmolarity of the RBCs = 300mOsM. 9 Osmolarity of the solution = 150mM x (2.0 osmoles/mole) = 300 mOsM Nothing. The RBCs will neither shrink nor swell B. Is the solution above hypotonic, hypertonic, or isotonic? Isotonic Exercise Red blood cells are placed in a beaker filled with 0.5L of solution containing 300mmoles of Urea. Urea is a penetrating solute and can cross the cell membrane. a) What will happen to the red blood cell? b) Is the solution above hypotonic, hypertonic, or isotonic? A. What will happen to the red blood cell? Osmolarity of the RBCs = 300mOsM. Osmolarity of the solution = (300mmole/0.5L) x (1.0 osmoles/mole) = 600 mOsM However, urea will cross the membrane until at equilibrium. The osmolarity of the solution after urea has moved to equilibrium will be 600 mOsM/2, or 300mOsM. 300mOsM will be added to the RBCs, making the osmolarity of the RBCs 300mOsM + 300mOsM, or 600mOsM. The RBCs will therefore swell and probably rupture. B. Is the solution above hypotonic, hypertonic, or isotonic? The solution is hypotonic because it causes the cells to swell and lyse, even though the solution is hyperosmotic Exercise How many grams would be needed to prepare 1L of a 5% glucose solution? Percent solution = the concentration of solutes in gram per 100 ml of aqueous solution. Accordingly, a 5% glucose solution would contain 5 grams per 100mL Because 1L (1000 ml) is needed, the amount of glucose would be (5g* 1000)/100=50g What is the molarity (M) of NaCl solution that contains 8.775 g per L? M=g per L + Molecular weight Molecular weight of NaCl 58.5, therefore M = 8.77 / 58.5 = 0.15 10 Exercise What is the osmolarity (osm) of a 0.1M CaCl2 solution? Osmolarity is the measure of osmotic pressure and is determined by number of particles One of molecule of CaCl2 is placed in solution would ionize and provides 3 particles (one Ca2+ and 2 Cl-) The osm (for molecules that ionize in solution = number ions from molecule * M = 3 * 0.1 M = 0.3 osm = 300 mOsm (milliosmole). Exercise How many grams would be required to make 1L of 300 mOsm NaCl solution? 300 mOsm NaCL = 150mM NaCl = 0.15 M NaCl g/L = M * MW = 0.15 * 58.5 = 8.775 g 11 Osmolarity and Molarity The molarity of a solution is equal to the number of moles of the specified substance that are dissolved in one liter of solution (1000 ml of solution). Therefore, we have in 1000 ml solution ...200 mM in 300 ml solution......x mM x =300*200/1000 =60 mM How Can I Prepare 3 Molar KCL Solution For Volume Of 500 Ml? The molar mass of KCl is 74.5 (39+35.5). This means that a 1molar solution would contain 74.5g KCl per liter. Therefore, a 3 molar solution would contain 74.5 x 3 = 223.5g KCl per liter. Now, 1 liter has 1000 ml in it. We need to prepare a 3M solution in 500 ml. Just simple math here. If 1000ml has 223.5g of KCl in it to make a 3M solution, 500ml will have half the amount (since 500 is the half of 1000). So, we'll divide 223.5 by 2. This gives us 111.75. So, first of all you need to weigh out 111.75g of KCl. You do this by first weighing the empty dish on the scale and calibrating the scale to zero with the dish on it. Then add KCl to the dish on the scale until the scale reads 111.75g. How would you prepare 300 mL of 0.750 M KCl? 16.7 grams of KCl diluted to a final volume of 300mL. Explanation: Molarity is equal to the number of moles of a solute per 1 liter of solution. (It's a way to measure concentration in chemistry). In this case, the KCl is the solute, and 0.750 M tells you how many moles of it we have per 1 liter of solution. Convert moles of KCl to grams using its molar mass, 74.5513 g/mol. 0.750mol x (74.5513g/1mol) = 55.883 grams. 1 In one liter of solution, there are 55.883 grams of KCl. But the question wants to know about 300mL (which is equal to 0.3L) 0.3L x 55.883g/1L = 16.7499 grams, round to 3 SF = 16.7 grams. If a solution is 0.5% (w/v), how many mg/ml is that? 5 mg mL−1 The solution's mass by volume percent concentration, m/v %, tells you the number of grams of solute present for every 100 mL of the solution. In your case, the solution is said to have a mass by volume percent concentration of 0.5%, which means that you get 0.5 g of solute for every 100 mL of the solution. 0.5% m/v=0.5 g solute per 100 mL of the solution You can thus say that 1 mL of this solution will contain 1mL solution *0.5 g solute/100mL solution =0.005 g solute To convert this to milligrams, use the fact that 1 g=103 mg You will end up with 0.005g*103 mg/1g = 5 mg Since this represents the number of milligrams of solute present for every 1 mL of the solution, you can say that the solution has a concentration of concentration = 5 mg mL−1. The answer is rounded to one significant figure. 2 Convert moles of a substance into grams A mole is the quantity of anything that has the same number of particles found in 12.000 grams of carbon-12. That number of particles is Avogadro's Number, which is roughly 6.02x1023. One mole of atoms contains 6 x 1023 atoms, no matter what element it is. In order to convert moles of a substance into grams you need a periodic table or another list of atomic masses. How many grams of carbon dioxide is 0.2 moles of CO2? Look up the atomic masses of carbon and oxygen. This is the number of grams per one mole of atoms. Carbon (C) has 12.01 grams per mole. Oxygen (O) has 16.00 grams per mole. One molecule of carbon dioxide contains 1 carbon atom and 2 oxygen atoms, so: number of grams per mole CO2 = 12.01 + [2 x 16.00] = 12.01 + 32.00 = 44.01 gram/mole Now, multiply this number of grams per mole (44.01 gram/mole) multiply by the number of moles (0.2 moles) in order to get the final answer: grams in 0.2 moles of CO2 = 0.2 moles x 44.01 grams/mole = 8.802 grams How to Convert Milligrams Per Liter to Molarity 1 Calculate the molecular weight of the solute in the solution. For example, assume the solute is sodium chloride, which has the molecular formula NaCl. To find the molecular weight of sodium chloride, add the weight of Na (22.99) to Cl (35.45), which equals 58.44 grams per mole. 2 To convert milligrams per liter to molarity, use this formula: Molarity = mg/L / (molecular weight of solute) x 1,000. Example: the solute in the solution is sodium chloride and the concentration is 0.5 mg/L, the equation would be Molarity = 0.5 mg/L / (58.44) x 1,000. The answer is 8.556 micromoles (μM). Convert from micromoles to moles or millimoles if necessary. 3 To calculate moles, use the equation Moles = micromoles / 1,000,000. To calculate millimoles, use the equation Millimoles = micromoles / 1,000 How to Interconvert Moles, Molarity and Volume 1 Calculate the molarity of a solution in moles per liter, given moles and volume in liters, by dividing the number of moles by the volume. For example, a 5.0 liter solution containing 10.0 moles has a molarity of 2.0 moles per liter. 2 Determine the number of moles in a solution, with molarity and volume known, by multiplying molarity in moles per liter by the volume in liters Example: a 2.0 liter solution with a molarity of 3.0 moles per liter. There are 6.0 moles in the solution. 3 Compute the volume of a solution in liters, given the number of moles and molarity, by dividing the number of moles by the molarity in units of moles per liter. Example: a solution containing 6.0 moles and a having a molarity of 3.0 moles per liter has a volume of 2.0 moles per liter. Exercise Calculate the number of grams of calcium hydroxide in 1.30 L of a 0.75 M Ca(OH)2 solution. Moles of Ca(OH)2 = 0.75 M x 1.30 L = 0.975 mol/L of Ca(OH)2 How many moles of potassium chloride (KCl) are contained in 0.25 L of 0.80 mol/L KCl? Moles of KCl = molarity x volume = 0.80 mol/L KCl x 0.25 L =0.20 mol/L Exercise A 4 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 ml teacup filled with hot water. What is the molarity of the sugar solution? M = m/V where M is molarity (mol/L) m = number of moles of solute V = volume of solvent (Liters) Determine number of moles of sucrose in 4 g C12H22O11 = (12)(12) + (1)(22) + (16)(11) 4 C12H22O11 = 144 + 22+ 176 C12H22O11 = 342 g/mol Divide this amount into the size of the sample 4 g/(342 g/mol) = 0.0117 mol Determine the volume of solvent in liters 350 ml x (1L/1000 ml) = 0.350 L Determine the molarity of the solution M = m/V M = 0.0117 mol /0.350 L M = 0.033 mol/L Answer: The molarity of the sugar solution is 0.033 mol/L. Exercise Calculate the Osmolarity of 0.3% Saline Each 100 mL contains 300 mg sodium chloride in water. This is 3 g per L. The solution is 3 g of NaCl dissolved in water to a total volume of 1000 mL. The molecular weight of NaCL is 58.5g per mole  58.5g NaCl = 1 mole 0.3% NaCl contains 3 g NaCl/1L solution The concentration is 3g/L divided by 58.5g/mole = 0.051 mole per liter.  Molarity of 3g of NaCI = 3g/L / 58.5g/mol = 0.051 mol/L OR 51 mmol/L solution of NaCl Since each molecule of NaCl dissociates into Na+ and Cl− ions Osmolarity of 0.3% saline  Molar value (51) is multiplied by 2 = 51*2 = 102 mOsm/L. Exercise Calculate the Osmolarity of 0.9% Saline Each 100 mL contains 900 mg sodium chloride in water. This is 9 g per L. The solution is 9 g of NaCl dissolved in water to a total volume of 1000 mL. The molecular weight of NaCL is 58.5g per mole  58.5g NaCl = 1 mole 0.9% NaCl contains 9 g NaCl/1L solution The concentration is 9g/L divided by 58.5g/mole = 0.154 mole per liter. 5  Molarity of 9g of NaCI = 9g/L / 58.43g/mol = 0.154 mol/L OR 154 mmol/L solution of NaCl Since each molecule of NaCl dissociates into Na+ and Cl− ions Osmolarity of 0.9% saline  Molar value (154) is multiplied by 2 = 154*2 = 308 mOsm/L. Exercise Find the osmolarity of 150 mM Sodium chloride (NaCl) 150 mM Sodium chloride (NaCl) has an osmolarity of  150 mM Na+ + 150 mM Cl− = 300 mOsmol.  150 mM NaCl has an osmolarity of 300 mOsmol. Find the osmolarity of 50 mM Calcium chloride (CaCl2) 50 mM Calcium chloride (CaCl2) and 5 mM Sodium bicarbonate (NaHCO3) have an osmolarity of  (1 * 50 mM Ca2+ + 2 * 50 mM Cl−) + (1 * 5 mM Na+ + 1 * 5 mM HCO3−) = 160 mOsmol.  50 mM CaCl2 and 5 mM NaHCO3 have an osmolarity of 160 mOsmol. Exercise If a red blood cell (typically 300 mOsmol/L) was placed into a solution of 100 mmol/L of NaCl (so 200 mOsmol/L because NaCL dissociates into 2 ions in a solution) then the cell would swell (the extracellular solution is hypotonic), If it was placed into a solution of 200 mmol/L of NaCl (400 mOsmol/L) then the cell would shrink (the solution is hypertonic), and if it was placed into a solution of 150 mmol/L of NaCl (300 mOsmol/L) the size of the cell would not change (the extracellular solution is isotonic). Exercise The osmolarity of a solution is the total number of solutes within the solution and if it is hypo/hyper/isosmotic it is in comparison to the osmolarity of another solution. Therefore, 100 mmol/L of NaCl would have an osmolarity of 200mOsmol/L. 6 100 mmol/L of NaCl plus 100 mmol/L of urea would have an osmolarity of 300 mOsmol/L because you add up ALL of the particles. NaCl makes up 2 ions in water  2 Osm/L  100mmol/L NaCl = 200mOsmol/L Urea does not dissociate in water  1 Osm/L  100mmol/L Urea = 100mOsmol/L The total osmolarity of the solution is NaCl 300 + Urea 100 = 300 mOsmol Exercise In order to determine the state of hyper/hypo/isosmolarity hyper/hypo/isotonicity of a solution you need to be comparing it TO something. OR Typically, we compare a solution to the state of a cell, which has 300mOsmol/L osmolarity Note that cells are made up of particles and proteins that CANNOT move across the plasma membrane. Therefore, to determine the tonicity (hyper / hypo / isotonicity) of a solution compared to the inside of a cell you would look at for the number of particles in that solution that cannot cross the plasma membrane (ex. ions) and compare that number to 300 mOsmol/L. So, 100 mmol/L of NaCl (200 mOsmol/L) would be hypotonic. It is also hypo-osmotic because it is only 200mOsm compared to 300mOsm of the cell The osmolarity of a solution containing both NaCl and Urea would therefore be 400 mOsmol/L. If we were to compare the osmolarity of this solution to that of a typical cell (300 mOsmol/L), then it would be hyperosmotic (the solution has an osmolarity of 400 mOsmol/L while the cell has an osmolarity of 300. If we were to compare the tonicity of this solution to that of a typical cell (300 mOsmol/L), then it would be isotonic (both have tonicities of 300 mOsmol/L). Therefore our solution is hyperosmotic and isotonic 7 Exercise We have two solutions (A and B) separated by a membrane that is permeable to water but not to solutes. Solution A was prepared by dissolving 5g of sodium chloride in 10L of water. Solution B was prepared by dissolving 12g of sodium chloride in 6L of water. a) What is the osmolarity of side A in the initial condition before osmotic equilibrium has been established? b) What is the osmolarity of side B in the initial condition before osmotic equilibrium has been established? c) Which direction will water flow to establish osmotic equilibrium in the condition above? (From A to B, from B to A, or there will be no net movement of water?) A. Molecular weight of sodium chloride = 58.44g/mole 1.0 mole NaCl = 2.0 osmoles (5g NaCl/10L) x (mol/58.44g) X (2.0 osmole/mole) = 0.017 osmoles B. Molecular weight of sodium chloride = 58.44g/mole 1.0 mole NaCl = 2.0 osmoles (12g NaCl/6L) x (mol/58.44g) X (2.0 osmole/mole) = 0.068 osmoles C. Water follows the solute, so from A to B Exercise Two solutions (A and B) are separated by a membrane that is permeable to water but not to solutes. Solution A contains 400osmoles of sodium chloride dispersed in a 5L solution. Solution B contains 240osmoles of sodium chloride dispersed in a 12L solution. a) What is the osmolarity of side A in the initial condition before osmotic equilibrium has been established? b) What is the osmolarity of side B in the initial condition before osmotic equilibrium has been established? c) Which direction will water flow to establish osmotic equilibrium in the condition above? (From A to B, from B to A, or there will be no net movement of water?) A. 400 osmoles/5L = 80 osmoles/L = 80 OsM B. 240 osmoles/12L = 20 osmoles/L = 20 OsM C. From B to A 8 Exercise A 5 liter solution with 20 osmoles of sucrose (side A) is separated from a 2 liter solution with 10 osmoles of sucrose (side B) by a membrane that is permeable to water but impermeable to sucrose. a) Which direction will water flow? (From A to B, from B to A, or there will be no net movement of water?) Side A: 20 osmoles/5 L = 4 OsM. Side B: 10 osmoles/2L = 5 OsM b) Which direction will water flow if the membrane is permeable to both sucrose in water? (From A to B, from B to A, or there will be no net movement of water?) A. From side A to B B. The solute will move to equilibrium, which means that both sides will have 4.5 OsM. Therefore, there will be no net movement of water. Exercise A 300 mOsM urea solution is ………… hyposmotic to a 200 mM solution of NaCl. (200mM NaCl = 400mOsM NaCl) Tonicity is …………… A relative term that describes the effect a solution has on cells of the body. True or False: A hypertonic solution will cause a cell to swell. False True or False: A solution can be hyperosmotic and still be hypo- or isotonic True Exercise Red blood cells are placed in a beaker filled with 0.5L of a 150mM NaCl solution. NaCl is a non-penetrating solute and cannot cross the red blood cell membrane. a) What will happen to the red blood cells? b) Is the solution above hypotonic, hypertonic, or isotonic? A. What will happen to the red blood cells? Osmolarity of the RBCs = 300mOsM. 9 Osmolarity of the solution = 150mM x (2.0 osmoles/mole) = 300 mOsM Nothing. The RBCs will neither shrink nor swell B. Is the solution above hypotonic, hypertonic, or isotonic? Isotonic Exercise Red blood cells are placed in a beaker filled with 0.5L of solution containing 300mmoles of Urea. Urea is a penetrating solute and can cross the cell membrane. a) What will happen to the red blood cell? b) Is the solution above hypotonic, hypertonic, or isotonic? A. What will happen to the red blood cell? Osmolarity of the RBCs = 300mOsM. Osmolarity of the solution = (300mmole/0.5L) x (1.0 osmoles/mole) = 600 mOsM However, urea will cross the membrane until at equilibrium. The osmolarity of the solution after urea has moved to equilibrium will be 600 mOsM/2, or 300mOsM. 300mOsM will be added to the RBCs, making the osmolarity of the RBCs 300mOsM + 300mOsM, or 600mOsM. The RBCs will therefore swell and probably rupture. B. Is the solution above hypotonic, hypertonic, or isotonic? The solution is hypotonic because it causes the cells to swell and lyse, even though the solution is hyperosmotic Exercise How many grams would be needed to prepare 1L of a 5% glucose solution? Percent solution = the concentration of solutes in gram per 100 ml of aqueous solution. Accordingly, a 5% glucose solution would contain 5 grams per 100mL Because 1L (1000 ml) is needed, the amount of glucose would be (5g* 1000)/100=50g What is the molarity (M) of NaCl solution that contains 8.775 g per L? M=g per L + Molecular weight Molecular weight of NaCl 58.5, therefore M = 8.77 / 58.5 = 0.15 10 Exercise What is the osmolarity (osm) of a 0.1M CaCl2 solution? Osmolarity is the measure of osmotic pressure and is determined by number of particles One of molecule of CaCl2 is placed in solution would ionize and provides 3 particles (one Ca2+ and 2 Cl-) The osm (for molecules that ionize in solution = number ions from molecule * M = 3 * 0.1 M = 0.3 osm = 300 mOsm (milliosmole). Exercise How many grams would be required to make 1L of 300 mOsm NaCl solution? 300 mOsm NaCL = 150mM NaCl = 0.15 M NaCl g/L = M * MW = 0.15 * 58.5 = 8.775 g 11 Percent by Weight Calculation Core Concepts In this tutorial, you will be introduced to a calculation called percent by weight. In order to fully understand, you will also walk through an example. What is Percent by Weight? Percent by weight, also referred to as mass percent composition, can be abbreviated as w/w%. This is a type of calculation that allows you to find the ratio of solute to solution. You can use the equation below to help solve example problems: How To Calculate Percent By Weight When doing this calculation, the way you arrive at your answer depends on the information given to you. However, each method ends the same way: you need to find the ratio of how many grams of solute there are for the total grams of solution. First, when doing any calculation, you have to make sure you are using consistent units. Sometimes the amounts in grams will be given, but sometimes you might have to do some extra work to get it to the same units. For example, if an amount is given in milligrams and the other in kilograms, you will have to convert using the correct conversions. You may also be given an amount in moles, which you can convert by using the molar mass. From now on, we will refer to the units in grams to stay consistent. Then, you have to find the total grams of solution, which includes the solvent and solute. You should already have the grams of solute, so you can use the same methods to find the grams of solvent. Lastly, you plug the numbers into the percent by weight equation and get your answer. Let’s walk through an example below to further understand this calculation! Percent by Weight Example Determine the percent by weight of sodium chloride in the following solution: 1.75 moles of NaCl (58.44 g/mol) dissolved in .550 kilograms of deionized H2O. To start this problem, we need to convert our information into the same unit, in this case, grams. Using molar masses and unit conversions, we are able to find the grams of solute (NaCl) and solvent (Water). 1 Since we have all the information we need, the final step is to plug everything into our formula. By doing this, we see that this solution is 15.68% NaCl. Calculating Percent Volume/Volume (% v/v) A percent v/v solution is calculated by the following formula using the milliliter as the base measure of volume (v): % v/v = mL of solute/100 mL of solution Example: What is the % v/v of a solution that has 5.0 mL of hydrochloric acid (HCl) diluted to 100 mL with deionized water? X % = 5.0 mL HCl/100 mL of solution X/100 = 5.0/100 Cross multiplying, 100X = 500 X = 5.0% % v/v Calculating Percent Weight/Volume (% w/v) A percent w/v solution is calculated with the following formula using the gram as the base measure of weight (w): % w/v = g of solute/100 mL of solution Example 1: Physiologic or isotonic saline is a 0.9% aqueous solution of NaCl. 2 0.9% saline = 0.9 g of NaCl diluted to 100 mL of deionized water, where NaCl is the solute and deionized water is the solvent. Example 1: What is the % w/v of a solution that has 7.5 g of sodium chloride diluted to 100 mL with deionized water? By definition, a percent w/v solution is the measure of weight per 100 mL. 7.5 g/100 mL = 7.5% You can calculate this value as well: X % = 7.5 g NaCl/100 mL of solution X/100 = 7.5/100 Cross multiplying, 100X = 750 X = 7.5% w/v What is the weight/volume percentage concentration of 250 mL of aqueous sodium chloride solution containing 5 g NaCl? 1. Write the equation: w/v% = w/v × 100 • weight/volume (%) = (mass solute ÷ volume of solution) × 100 2. Identify the solute and solvent (by name or chemical formula) • solute = sodium chloride = NaCl • solvent is water (H O). 2 3. Extract the data from the question (mass of solute, volume of solution) • mass solute (NaCl) = 5 g and volume of solution = 250 mL 4. Check the units for consistency and convert if necessary (mass in grams, volume in millilitres) • mass solute (NaCl) = 5 g (no unit conversion needed) • volume of solution = 250 mL (no unit conversion needed) 5. Substitute these values into the equation and solve. • w/v (%) = (5 g ÷ 250 mL) × 100 = 2 g/100 mL How do you convert from %w/v to molarity? We define %w/v as: By unit conversion, we have that and that Define: • msolute for the solute mass in g • Vsoln for the solution volume in mL • Msolute for the molar mass of the solute in g/mol 3 Therefore: Or, rewriting in terms of %w/v and implied unit cancellation, we have: As an example, if we have a 37% w/v aqueous HCl solution, then: 4 Weight/Volume Percentage Concentration (w/v %) Key Concepts • • • • • • Weight/Volume Percentage Concentration is a measurement of the concentration of a solution. weight/volume percentage concentration is usually abbreviated as w/v (%) To calculate w/v % concentration: mass of solute (g) w/v (%) = × 100 volume of solution (mL) Common units(1) for w/v% concentration are g/100 mL : (i) Solubilities are sometimes given in units of grams of solute per 100 mL of aqueous solution, that is, as a weig

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