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The solubility of $N_{2}$ , in water at 300 Kand 500 Torr partial pressure is $0.01 {gL}^{- 1}$ • The solubility (in ${gL}^{- 1})$ at 750 Torr partial pressure is

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0.0075

0.005

0.02

0.015

answer is D.

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Detailed Solution

According to Henry's law, $s = k_{H} p$ . Hence $\frac{s_{2}}{s_{1}} = \frac{p_{2}}{p_{1}} i.e. s_{2} = (\frac{p_{2}}{p_{1}}) s_{1} = (\frac{750 Torr}{500 Torr}) (0.01 {gL}^{- 1}) = 0.015 {gL}^{- 1}$

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