Question

$H_{2}S$, a toxic gas with a rotten egg-like smell, is used for the qualitative analysis. If the solubility of $H_{2}S$ in the water at STP is $0.195$ m, then Henry's law constant (in atm) is :

• A. $28.94$
• B. $285$
• C. $145.2$
• D. $2890.4$

Answer 1

Answer: (B)

$285$

According to Raoult's Law, the vapour pressure of a volatile component in a given solution is given by,

$P_i=X_i{P}_i^o$

and according to Henry's Law, when one of the components is so volatile that it exists as a gas,

$p=K_{H}x$

$K_H=$ Henry's constant, $x=$ mol fraction

At STP, $p=1atm$

Mol fraction of $H_{2}S=\frac{0.195}{0.195+55.5}=0.0035=x$

$\Rightarrow p=K_H\times x\Rightarrow K_H=\frac{1.00}{0.0035}=285.7atm$

Hence the correct option is B.