Question

Molality of a sulphuric acid solution in which the mole fraction of water is $0.85$ is :-

• A. $4.9$
• B. $9.8$
• C. $19.6$
• D. $0.15$

Answer 1

Answer: (B)

$X_{H_{2}O}=0.85X_{H_{2}S{O}_4}=0.15$
$\frac{X_{H_{2}S{O}_4}}{X_{H_{2}O}}=\frac{m_{H_{2}S{O}_4}}{m_{H_{2}O}}=\frac{0.15}{0.85}$
$m=\frac{0.15\times 1000}{0.85\times 18}=9.8m$