Question

Calculate the solubility of certain gas in solvent with pressure \(3\) atm at \(25^{\circ}\text{C}\) (Henry's law constant is \(3.0\times 10^{-2}\,\text{mol dm}^{-3}\,\text{atm}^{-1}\))

• A. \(0.07\) M
• B. \(0.08\) M
• C. \(0.09\) M
• D. \(0.1\) M

Hint:

Henry law: solubility = KH x pressure.

Answer 1

The Correct Option is C

Step 1: Understand the law
Henry's law says the solubility of a gas in a liquid is directly proportional to its partial pressure: \(S = K_H \times P\).

Step 2: Substitute the data
\(K_H = 3.0 \times 10^{-2}\) mol dm\(^{-3}\) atm\(^{-1}\) and \(P = 3\) atm.

Step 3: Calculate
\[ S = 3.0 \times 10^{-2} \times 3 = 9.0 \times 10^{-2} = 0.09\ \text{mol dm}^{-3} \]

Step 4: Check the options
The values 0.07, 0.08 and 0.1 M would need pressures of 2.33, 2.67 and 3.33 atm. Only 0.09 M corresponds to exactly 3 atm.

Final Answer:
The gas solubility is 0.09 M. This is option (C). \[ \boxed{\text{(C) }0.09\ \text{M}} \]