Question

The solubility of \(\text{N}_2\) gas in water at \(25^{\circ}\)C and \(1\) bar is \(6.85\times 10^{-4}\text{ mol L}^{-1}\). Calculate the solubility of \(\text{N}_2\) gas in water at the same temperature when the partial pressure of \(\text{N}_2\) is \(0.70\) bar

• A. \(5.138\times 10^{-4}\text{ mol L}^{-1}\)
• B. \(5.480\times 10^{-4}\text{ mol L}^{-1}\)
• C. \(4.795\times 10^{-4}\text{ mol L}^{-1}\)
• D. \(4.875\times 10^{-4}\text{ mol L}^{-1}\)

Hint:

Henry's law: solubility is directly proportional to partial pressure.

Answer 1

The Correct Option is C

Step 1: Use the Henry constant
$K_H=\frac{p}{S}$ stays fixed. $\frac{S_2}{S_1}=\frac{p_2}{p_1}=0.70$.
So $S_2=0.70\times6.85\times10^{-4}=4.795\times10^{-4}$, option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]