Question

Find the molality of a solution made by mixing equal volumes of 60% (w/w) $H_{2}S{O}_4$ (density = 1.84 g/mL) and 40% (w/w) of $H_{2}S{O}_4$ (density = 1.52 g/mL) solutions.

• A. 10.6 m
• B. 12.5 m
• C. 19.2 m
• D. 15.7 m

Answer 1

The correct option is A 10.6 m

Assuming V mL of $H_{2}S{O}_4$ is mixed together.
60% (w/w) means 60 g of solute is present in 100 g of the solution.
Since, density of solution = $\frac{\text{mass of the solution}}{\text{volume of the solution}}$
Mass of $H_{2}S{O}_4$ solution = 1.84 $\times V$ = 1.84V g
100 g of the solution contains 60 g of $H_{2}S{O}_4$
1.84V g of the solution will contain = $\frac{60}{100}\times 1.84V$ = 1.104V g...(i)
Similarly,
40% (w/w) means 40 g of solute is present in 100 g of the solution.
Since, density = $\frac{\text{mass of the solution}}{\text{volume of the solution}}$
Mass of $H_{2}S{O}_4$ solution = 1.52 $\times V$ = 1.52V g
100 g of the solution contains 40 g of $H_{2}S{O}_4$
1.52V g of the solution will contain = $\frac{40}{100}\times 1.52V$ = 0.608V g...(ii)
Total mass of the solution = 1.84V + 1.52V = 3.36V g
Total mass of the solute = 1.104V + 0.608V = 1.712V g
Moles of $H_{2}S{O}_4$ = $\frac{\text{given mass}}{\text{molar mass}}=\frac{1.712V}{98}$ mol
Total mass of the solvent (water) = 3.36V - 1.712V = 1.648V g
Molarity = $\frac{\text{moles of solute}}{\text{mass of solution in kg}}$
Molality = $\frac{1.712V/98}{1.648V}\times 1000$ = 10.6 m