Question

A solution is made by mixing equal volumes of $30\%$ by weight of $H_{2}S{O}_4$ (density = $1.218$ g/mL) and $70\%$ by weight of $H_{2}S{O}_4$ (density = $1.610$ g/mL).
Then which of the following statement(s) is/are correct?

• A. Weight of $H_{2}O$ (solvent) is $133.56g$
• B. Molarity of the final solution is $7.6M$
• C. Molality of the final solution is $11.4m$
• D. Molarity of the final solution is $5.4M$

Answer 1

Answer: (A), (B), (C)

Molality of the final solution is $11.4m$

Suppose that the solution contains $100$ mL of each variety of $H_{2}S{O}_4.$ Total volume is, therefore, $200$ mL or $0.2$ L
Weight of $100$ mL of $H_{2}S{O}_4$ solution $\left(30\%\right)=1.218\times 100$
$=121.8g$
and weight of $100$ mL of $H_{2}S{O}_4$ solution $\left(70\%\right)$
$=1.610\times 100$
$=161g$
$\text{Weight of}{H}_{2}S{O}_4\left(30\%\right)=121.8\times \frac{30}{100}=36.54g$
$\text{Weight of}{H}_{2}S{O}_4\left(70\%\right)=161\times \frac{70}{100}=112.7g$
Total weight of $H_{2}S{O}_4$ (solute)
$=36.54+112.7=149.24$ g
$\therefore$ Weight of $H_{2}O$ (solvent) = weight of solution - weight of solute
$=(121.8+161)-149.24$
$=133.56g$
$\text{Moles of}$ $H_{2}S{O}_4=\frac{149.24}{98}=1.5228$ (molecular weight of $H_{2}S{O}_4=98)$
$\text{Molality}=\frac{1.5228}{133.56}\times 1000=11.4m$

$\text{Molarity}=\frac{1.5228}{0.2}=7.6M$