Answer
$\text { (d) } 0.13 \mathrm{~kg} / \mathrm{m}^3$
Work Step by Step
Recall Henry’s law in concentration form: $$C=\frac{p}{H}.$$ We are given the solubility $S$: $$S=\frac{1}{H},$$ so $$C=S\cdot p.$$ We have: $$H=\frac{1}{S}=\frac{1}{0.00156}\approx 641\text{ bar} \cdot \text{m}^3\text{/kmol}.$$ At $p=3\text{ bar}$ we have: $$C=\frac{3}{641}\approx 0.00468\text{ kmol/m}^3.$$ We convert to mass: $$\rho=0.00468\times 28\approx 0.131\text{ kg/m}^3.$$ The correct answer is (d).
