The partial pressure of ethane over a solution containing

Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1

= 30 g mol−1

$\therefore$ Number of moles present in $6.56 \times 10^{-3} \mathrm{~g}$ of ethane $=\frac{6.56 \times 10^{-3}}{30}$

= 2.187 × 10−4 mol

Let the number of moles of the solvent be x.

According to Henry’s law,

p = KHx

 $\Rightarrow 1$ bar $=K_{H} \cdot \frac{2.187 \times 10^{-4}}{2.187 \times 10^{-4}+x}$

$\Rightarrow 1$ bar $=K_{H} \cdot \frac{2.187 \times 10^{-4}}{x}$ (Since $\left.x>>2.187 \times 10^{-4}\right)$

$\Rightarrow K_{H}=\frac{x}{2.187 \times 10^{-4}}$ bar

Number of moles present in $5.00 \times 10^{-2} \mathrm{~g}$ of ethane $=\frac{5.00 \times 10^{-2}}{30} \mathrm{~mol}$

= 1.67 × 10−3 mol

According to Henry’s law,

p = KHx

$=\frac{x}{2.187 \times 10^{-4}} \times \frac{1.67 \times 10^{-3}}{\left(1.67 \times 10^{-3}+x\right.}$

$=\frac{x}{2.187 \times 10^{-4}} \times \frac{1.67 \times 10^{-3}}{x} \quad\left(\sin c e x>>1.67 \times 10^{-3}\right)$

= 7.636 bar

Hence, partial pressure of the gas shall be 7.636 bar.