Question

${\mathrm{H}}_2\mathrm{S}$, toxic gas with a rotten egg-like smell, is used for the qualitative analysis. If the solubility of ${\mathrm{H}}_2\mathrm{S}$ in water at STP is 0.195 m. Calculate Henry's law constant.

• A. 213.8 bar
• B. 282.0 bar
• C. 345.8 bar
• D. 462.9 bar

Answer 1

The correct option is B

282.0 bar

The explanation for the correct answer

Calculation of Henry's law constant

• According to Henry's law:
  ${\mathrm{x}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{H}}\times {\mathrm{p}}_{\mathrm{A}}$
• Here, ${\mathrm{x}}_{\mathrm{A}}$is the mole fraction of solute, ${\mathrm{K}}_{\mathrm{H}}$ is Henry's law constant, and ${\mathrm{p}}_{\mathrm{A}}$ is partial pressure.

Step 1: Given data

• Given the solubility of ${\mathrm{H}}_2\mathrm{S}$ in water = 0.195 m at STP(Standard Temperature and Pressure)
• Here, the solution is of ${\mathrm{H}}_2\mathrm{S}$ and water, where the solute is ${\mathrm{H}}_2\mathrm{S}$ and the solvent is water.

Step 2: Calculation of the number of moles
Number of moles in 1 kg of solvent = $\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}$
The molar mass of ${\mathrm{H}}_2\mathrm{O}$ = $2\times 1+16$
The molar mass of ${\mathrm{H}}_2\mathrm{O}$ = 18
Number of moles of solvent = $\frac{1000}{18}$
Number of moles of solvent = 55.55 moles.

Step 3: Calculation of mole fraction of ${\mathrm{H}}_2\mathrm{S}$
Mole fraction of ${\mathrm{H}}_2\mathrm{S}$ = $\frac{0.195}{0.195+55.55}$
Mole fraction of ${\mathrm{H}}_2\mathrm{S}$ = $\frac{0.195}{55.745}$
Mole fraction of ${\mathrm{H}}_2\mathrm{S}$ = 0.0035

Step 4: Calculation of Henry's law constant
Pressure at STP = 0.987 bar
Henry's law constant for ${\mathrm{H}}_2\mathrm{S}$ = $\frac{0.987}{0.0035}$

Therefore, Henry's law constant for ${\mathrm{H}}_2\mathrm{S}$ = 282 bar, option (B) is correct.