Question

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

• A. $300$ bar
• B. $282$ bar
• C. $100$ bar
• D. $125$ bar

Answer 1

Answer: (B)

Solubility of $H_{2}S$ gas $=0.195m$
$\therefore$ Moles of $H_{2}S=0.195$, Mass of water $=1000g$
No. of moles of water $=\frac{1000g}{18gmo{l}^{-1}}=55.55$ moles
$\therefore$ Mole fraction of $H_{2}S$ gas in the solution, (x)
$=\frac{0.195}{0.195+55.55}=\frac{0.195}{55.745}=0.0035$
Pressure at $STP=0.987$ bar
Applying Henry’s law, $p(H_{2}S)=K_H\times x_{H_{2}S}$
or $K_H=\frac{P_{H_{2}S}}{x_{H_{2}S}}=\frac{0.987bar}{0.0035}=282$ bar