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: State Henry's law
The solubility of a gas is directly proportional to its partial pressure: \(S\propto p\).

Step 2: Compute
\[ S_2=6.85\times10^{-4}\times\frac{0.70}{1}=4.795\times10^{-4}\text{ mol L}^{-1} \]

Step 3: Result
Option (C).

Final Answer:
Solubility is \(4.795\times10^{-4}\) mol L\(^{-1}\), option (C). \[ \boxed{\text{(C)}} \]