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 :

Question

Which one of the following is not correct for an ideal solution?

Answer 1

To solve the problem regarding the solubility of nitrogen ($N_2$) in water, we need to apply Henry's Law. Henry's Law states that the solubility of a gas in a liquid at a fixed temperature is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it is expressed as:

S = k_H \cdot P

where:

S is the solubility of the gas in the liquid (in g/L in this context),
k_H is Henry's law constant, and
P is the partial pressure of the gas (in atm or other consistent units).

Given:

At 300 K and 500 torr, the solubility of $N_2$ is 0.01 g/L.
We need to find the solubility at 750 torr.

From Henry's Law, since the temperature remains constant, the ratio of solubility to pressure should remain constant. Therefore, we can write:

\frac{S_1}{P_1} = \frac{S_2}{P_2}

Substituting the given values:

S_1 = 0.01 \, \text{g/L}
P_1 = 500 \, \text{torr}
P_2 = 750 \, \text{torr}

We need to find S_2, the solubility at 750 torr:

\frac{0.01}{500} = \frac{S_2}{750}

Solving for S_2:

S_2 = 0.01 \cdot \frac{750}{500}

S_2 = 0.01 \cdot 1.5

S_2 = 0.015 \, \text{g/L}

Therefore, the solubility of $N_2$ at 750 torr is 0.015 g/L, which matches the correct answer option 0.015.

Answer 2