Ncert Exercises & Solutions

2.6 H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.

It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.

Moles of water $= \frac{1000 g}{18 g {mol}^{- 1}} = 55.56 mol$

Mole fraction of H₂S, $x = \frac{Moles of H_{2} S}{Moles of H_{2} S + Moles of water}$

$= \frac{0.195}{0.195 + 55.56}$

=0.0035

At STP, pressure (p) = 0.987 bar

According to Henry’s law:

p = K_Hx

$K_{H} = \frac{p}{x}$
$= \frac{0.987}{0.0035} bar$

= 282 bar

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