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  4. The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L-1. The solubility (in g L-1) at 750 torr partial pressure is :

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Q. The solubility of $N_2$ in water at $300 \,K$ and $500$ torr partial pressure is $0.01 \,g \,L^{-1}$. The solubility (in g $L^{-1}$) at $750$ torr partial pressure is :

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Solution:

Partial Pressure = Mole fraction × solubility
$\frac{p_{1}}{p_{2}}=\frac{s_{1}}{s_{2}} \Rightarrow \frac{500}{0.01}=\frac{750}{x}$
$\therefore x=0.015\,g / L$