Question

Henry's law constant for solubility of methane in benzene is $4.27\times {10}^{-5}$ $m{m}^{-1}Hgmol$ $d{m}^{-3}$ at constant temperature. Calculate solubility of methane at $760mmHg$ pressure at same temperature.

Answer 1

We are given, pressure p = 760 mm Hg

$K_H=4.27\times {10}^{-5}m{m}^{-1}$ Hg mol $d{m}^{-3}$

According to Henry's law:

$p=K_{H}x$ (where x is the solubility of methane)

On substituting values, we get:

$760=4.27\times {10}^{-5}\times x$

or $x=\frac{760}{4.27\times {10}^{-5}}=1.77\times {10}^7$