Question

Fig.

If the initial pressure of a gas 0.03 atm, the mass of the gas absorbed per gram of the adsorbent is ____× 10–2g

Answer 1

The correct answer is 12
Given that log K = intercept = 0.602 = log 4
∴ K = 4
Slope $=\frac{1}{n}=1$ and
initial pressure = 0.03 atm
$\frac{x}{m}=K(p)^{\frac{1}{n}}$
= 4 × 0.03 = 0.12 = 12 × 10–2
∴ mass of gas absorbed per gm of adsorbent = 12 × 10–2 g