Question

H2S, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195m, calculate Henrys law constant.

Answer 1

given, Molality =0.195 m,

Let take mass of solvent (water) = 1 kg

Use above formula we get

Moles of solute = 0.195 mol

Molar mass of water (H2O ) = 2 × 1 + 16 = 18

Mass of water (mass of solvent 1 kg = 1000 g

Number of moles of water = 1000 g / 18 = 55.56 mol

Use above formula we get

Mole fraction of H2S = 0.195/(0.195 + 55.56) = 0.0035

At STP, pressure (p) = 0.987 bar always

According to Henry’s law:

p = KH × X

KH = p / X = 0.987 / 0.0035 = 282 bar