Biochemistry Lab Solution Preparation Quiz

<p>100 ml</p> Signup and view all the answers

<p>16 g</p> Signup and view all the answers

<p>The molecular weight of NaCl is calculated by adding the molar masses of sodium (Na) and chlorine (Cl): $22.99 , \text{g/mol} (Na) + 35.45 , \text{g/mol} (Cl) = 58.44 , \text{g/mol}$.</p> Signup and view all the answers

<p>To calculate the moles of NaCl needed, we can use the formula: Moles = Normality $\times$ Volume (in liters). So, for a 0.1 N solution and 100 ml (0.1 L) volume, the moles of NaCl needed would be: Moles = 0.1 N $\times$ 0.1 L = 0.01 moles.</p> Signup and view all the answers

<p>In this context, normality (N) is defined as the number of equivalents of a solute per liter of solution.</p> Signup and view all the answers

<p>To calculate the mass of NaCl required, we can use the formula: Mass = Moles $\times$ Molecular weight. Using the previously calculated moles (0.01) and the molecular weight of NaCl (58.44 g/mol), the mass of NaCl required would be: Mass = 0.01 moles $\times$ 58.44 g/mol = 0.5844 grams.</p> Signup and view all the answers

<p>To calculate the equivalents of NaCl, we can use the formula: Equivalents = Normality $\times$ Volume (in liters). For a 0.1 N solution and 100 ml (0.1 L) volume, the equivalents of NaCl present would be: Equivalents = 0.1 N $\times$ 0.1 L = 0.01 equivalents.</p> Signup and view all the answers

<p>The molecular weight of NaCl is calculated by adding the molar masses of sodium (Na) and chlorine (Cl): [ \text{Molecular weight of NaCl} = 22.99 , \text{g/mol (Na)} + 35.45 , \text{g/mol (Cl)} = 58.44 , \text{g/mol} ]</p> Signup and view all the answers

<p>To prepare a 0.1 N solution of NaCl, the number of moles needed can be calculated using the formula: [ \text{Number of moles} = \text{Normality} \times \text{Volume (in liters)} = 0.1 \times 0.1 = 0.01 , \text{moles} ]</p> Signup and view all the answers

<p>In this context, normality (N) is defined as the number of equivalents of a solute per liter of solution. For NaCl, as it is a simple ionic compound, 1 mole of NaCl is equivalent to 1 mole of Na+ ions and 1 mole of Cl- ions.</p> Signup and view all the answers

<p>The volume of solution containing 1 mole of NaCl for a 0.2 N solution can be calculated using the formula: [ \text{Volume (in liters)} = \frac{\text{Number of moles}},{\text{Normality}} = \frac{1},{0.2} = 5 , \text{liters} ]</p> Signup and view all the answers

<p>The grams needed to prepare a 0.2 N solution can be calculated using the formula: [ \text{Grams} = \text{Number of moles} \times \text{Molar mass} = 0.1 \times 40 = 4 , \text{grams} ]</p> Signup and view all the answers