Question

The partial pressure of a gas at \(25^{\circ}\)C is \(0.18\) atm, calculate the concentration of the gas dissolved at the same temperature,
If \(K_H\) is \(0.15 \text{mol dm}^{-3} \text{atm}^{-1}\)

• A. \(0.027\) M
• B. \(1.8\) M
• C. \(4.5\) M
• D. \(0.45\) M

Hint:

Henry law: concentration = KH x partial pressure.

Answer 1

The Correct Option is A

Step 1: Use the proportionality:
Doubling the pressure doubles the dissolved gas. The constant $K_H$ tells how many mol/dm$^3$ dissolve per atm.

Step 2: Multiply:
$0.15 \times 0.18$: first $15 \times 18 = 270$. The two numbers have 2 and 2 decimal places, so the result is $270 \times 10^{-4} = 0.027$.

Step 3: Check:
The unit works: (mol dm$^{-3}$ atm$^{-1}$) times atm = mol dm$^{-3}$. Option A is right.

Final Answer:
Multiplying KH by the pressure gives 0.027 M. \[ \boxed{\text{(A) }0.027\ \text{M}} \]