Question

$H_{2}S$ a toxic gas with rottenegg like smell is used for the qualetative analysis, if the solubility of $H_{2}S$ in water,at STP is $0.195mole.k{g}^{-1}$, the Henry's law constant is:

• A. 282 bar
• B. 324.8 bar
• C. 462.9 bar
• D. 534.8 bar

Answer 1

Answer: (A)

282 bar

Solubility of $H_{2}S=0.195$ moles/kg

$1$ kg of solvent contains $=\frac{1000}{18}=55.55$ moles

$\therefore$ $x_{H_{2}S}=\frac{n_{H_{2}S}}{n_{H_{2}S}+n_{H_{2}O}}=\frac{0.195}{0.195+55.55}=\frac{0.195}{55.745}=0.0035$

Pressure at $STP=0.987$ bar $=1$ atm

$P_{H_{2}S}=K_H\times x_{H_{2}S}$

$\therefore$ $K_H=\frac{0.987bar}{0.0035}=282$ bar

Hence, the correct option is $A$