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 :

Question

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 : (A) 0.0075 (B) 0.015 (C) 0.02 (D) 0.005

Tardigrade Admin · Accepted Answer

Partial Pressure = Mole fraction Ã solubility (p1/p2)=(s1/s2) ⇒ (500/0.01)=(750/x) ∴ x=0.015 g / L

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 :

0.0075

0.015

0.02

0.005

Partial Pressure = Mole fraction × solubility
$\frac{p _{1}}{p _{2}} = \frac{s _{1}}{s _{2}} \Rightarrow \frac{500}{0.01} = \frac{750}{x}$
$∴ x = 0.015 g / L$

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