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 $0.195m$, then Henry's law constant (in atm) is nearly :

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

Answer 1

The correct option is B 285

According to Henry's Law,
$P=K_H\times \chi$
$K_H=\text{Henry's constant},\chi =\text{mol fraction}$
At STP, P = 1 atm
$0.195m$ means 0.195 mol of gas is dissolved in 1 kg of solvent (water)
$n_{H_{2}O}=\frac{1000g}{18gmo{l}^{-1}}$

$\text{Mol fraction of}{H}_{2}S=\frac{n_{H_{2}S}}{n_{H_{2}S}+n_{H_{2}O}}$

${\chi }_{H_{2}S}=\frac{0.195}{0.195+\frac{1000}{18}}=0.0035\Rightarrow P=K_H\times \chi \Rightarrow K_H=\frac{1.00}{0.0035}\approx 285.7atm$