Solved Problems Lab 2 PDF

Solutions Concentration Units Concentration is the amount of solute present in a given amount of solution. Moles of solute Molarity (M)= Liters of solution Molarity has the units of...

Solutions Concentration Units Concentration is the amount of solute present in a given amount of solution. Moles of solute Molarity (M)= Liters of solution Molarity has the units of mole per liter (mol/L).  Solutions Mass of solute Percent by mass of solute = x 100 % mass solute + mass solvent Mass of solute = x 100% Mass of soln. The percent by mass is a unitless number because it is a ratio of two similar quantities.  Solutions The mole fraction of a component of :the solution Moles of A mole fraction of component A = XA = Sum of moles of all components The Molality (m) : Moles of solute molality = Mass of solvent (kilograms) Molality is the number of moles of solute dissolved in 1 kg of solvent.  Solutions EXAMPLE 2: a. What is the percent of MgSO4 by mass in a solution made from 16.0g MgSO4 (M.wt. 120.4 g) and 100 mL of H2O (M.wt. 18 g) at 25oC ? The density of water at 25oC is 0.997 g/ml b. What is the mole fraction of each component. c. Calculate the molality of the solution.  Solutions a. What is the percent of MgSO4  Solutions b. What is the mole fraction of each component. =1  Solutions c. Calculate the molality of the solution  Solutions EXAMPLE 3: Calculate the molality of a sulfuric acid solution containing 35.2 g of sulfuric acid in 237 g of water. The molar mass of sulfuric acid is 98.09 g. Molality = mol of H2SO4 / Kilograms of H2O mol of H2SO4 = mass / molar mass = 35.2 / 98.09 = 0.358 mol Kilograms of H2O = 237 / 1000 = 0.237 Kg Molality = 0.358 / 0.237 = 1.5 m  Solutions EXAMPLE 4: What is the molality of a solution containing 7.78 g of urea [(NH2)2CO] in 203 g of water? The molar mass of urea is 60.06 g Molality = mol of urea / Kilograms of H2O mol of urea = mass / molar mass = 7.78 / 60.06 = 0.129 mol Kilograms of H2O = 203 / 1000 = 0.203 Kg Molality = 0.129 / 0.203 = 0.638 m  Solutions EXAMPLE 5: Calculate the molality of a 29.7 percent (by mass) aqueous solution of phosphoric acid (H3PO4). The molar mass of phosphoric acid is 97.99 g. Mass of H3PO4 = 29.7 g Mass of H2O = 70.3 g Molality = mol of H3PO4 / Kilograms of H2O mol of H3PO4 = mass / molar mass = 29.7 / 97.99 = 0.303 mol Kilograms of H2O = 70.3 / 1000 = 0.07 Kg Molality = 0.303 / 0.07 = 4.32 m  Solutions EXAMPLE 6: Calculate the molality of a 44.6 percent (by mass) aqueous solution of sodium chloride. The molar mass of sodium chloride is 60.06 g Mass of NaCl = 44.6 g Mass of H2O = 55.4 g Molality = mol of NaCl / Kilograms of H2O mol of NaCl = mass / molar mass = 44.6 / 60.06 = 0.742 mol Kilograms of H2O = 55.4 / 1000 = 0.055 Kg Molality = 0.742 / 0.055 = 13.5 m  Solutions EXAMPLE 7: Concentrated hydrochloric acid is 36.5 percent HCl by mass. Its density is l.189 g/ml. a. Calculate the molality.  Solutions EXAMPLE 7: Concentrated hydrochloric acid is 36.5 percent HCl by mass. Its density is l.189 g/ml. b. Calculate the molarity of concentrated HCl.  How would you prepare 425 gm of an aqueous solution containing 2.40% by mass of sodium acetate, NaC2H3O2? Problem Strategy In order to know how to prepare this solution, we need to know the mass of solute (NaC2H3O2) and the mass of solution. The mass percent given in the problem (2.40%) will enable us to determine the mass of solute present in the 425 gm sample. Keeping in mind that the total mass of solution equals the sum of the mass of solute and the mass of solvent (mass solution = mass solute + mass solvent), we will be able to determine the mass of H2O needed to prepare the solution. Solution The mass of sodium acetate (solute) in 425 g of solution is Mass of NaC2H3O2 = 425 g × 0.0240 = 10.2 g We have the relationship Mass of solution = mass NaC2H3O2 + mass H2O Therefore, the quantity of water in the solution is Mass of H2O = mass of solution − mass of NaC2H3O2 = 425 g − 10.2 g = 415 g You would prepare the solution by dissolving 10.2 g of sodium acetate in 415 g of water.  Exercise An experiment calls for 35.0 g of hydrochloric acid that is 20.2% HCl by mass. How many grams of HCl is this? How many grams of water?  Henry’s law Relates gas concentration c, in moles per liter, to the gas pressure p in atmospheres cαP c=kP where k is a constant for each gas, and has units of mol/L atm. Example: A liter of water at 25°C dissolves 0.0404 g O2 when the partial pressure of the oxygen is 1.00 atm. What is the solubility of oxygen from air, in which the partial pressure of O 2 is 159 mmHg?