Question 13.1.2: A solution consists of 20.4 g of silver nitrate, AgNO3, and ......
A solution consists of 20.4 g of silver nitrate, AgNO_{3}, and 113.6 g of water.
a. Calculate the weight percent, the molality, and the mole fraction of AgNO_{3} in the solution.
b. The solution volume is 117 mL. Calculate the molarity of AgNO_{3} in the solution
You are asked to calculate the concentration of a solution in weight percent, molality, mole fraction, and molarity units.
You are given the mass of solute and solvent in the solution and the total volume of the solution.
Convert the quantities of solute and solvent to units of moles.
20.4 g AgNO_{3}\times \frac{1\text{ mol AgNO}_{3}}{169.9 \text{ g}} = 0.120 mol AgNO_{3}
113.6 g H_{2}O \times \frac{1\text{ mol H}_{2}\text{O}}{18.015 \text{ g}} = 6.306 mol H_{2}O
a. Use the definition of each concentration unit to calculate the solution concentration.
% AgNO_{3}=\frac{20.4 \text{ g AgNO}_{3}}{20.4\text{ g AgNO}_{3}+113.6 \text{ g H}_{2}\text{O}} \times 100% = 15.2%
molality AgNO_{3}=\frac{0.120\text{ mol AgNO}_{3}}{(113.6 \text{ g H}_{2}\text{O})\left(\frac{1\text{ kg}}{10^{3}\text{ g}} \right) } = 1.06 m
χ_{AgNO_{3}}=\frac{0.120\text{ mol AgNO}_{3}}{0.120\text{ mol AgNO}_{3}+6.30\text{ mol H}_{2}\text{O}} = 0.0187
a. Use the definition of molarity to calculate the solution concentration.
molarity AgNO_{3}=\frac{0.120\text{ mol AgNO}_{3}}{(117 \text{ mL})\left(\frac{1\text { L}}{10^{3}\text{ mL}} \right) } = 1.03 M