Question

What is the molality of solution containing $24.5\%$ of $H_{2}S{O}_4$ by mass? The density of solution is $1.093$ g/mL

• A. $2.90$ m
• B. $2.73$ m
• C. $3.01$ m
• D. $3.31$ m

Answer 1

The correct option is D $3.31$ m

$24.5$% of $H_{2}S{O}_4=24.5gm$ of $H_{2}S{O}_4$ in $100gm$ of $H_{2}O$ .

Number of moles of $H_{2}S{O}_4=\frac{24.5}{98}=0.25$ mol

Weight of solvent= $100-24.5=75.5\times {10}^{-3}kg$ .

Molality= $\frac{0.25}{75.5\times {10}^{-3}}=3.31m$ .