Question

Find the molality of $H_{2}S{O}_4$ solution whose specific gravity is $1.98gm{l}^{-1}$ and 95% by volume $H_{2}S{O}_4$

• A. $7.412$
• B. $8.412$
• C. $9.412$
• D. $10.412$

Answer 1

Answer: (C)

$H_{2}S{O}_4$ is 95% by volume
wt. of $H_{2}S{O}_4=95g$
Vol of solution = 100 ml
$\therefore$ Moles of $H_{2}S{O}_4=\frac{95}{98}$, and weight of solution
$=100\times 1.98=198g$
Weight of water $=198-95=103g$
Molality $=\frac{95\times 1000}{98\times 103}=9.412$
Hence molality of $H_{2}S{O}_4$ solution is 9.412