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