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Standard XII

Chemistry

Introduction

When a graph ...

Question

When a graph is plotted between log x/m and log P, it is straight line with an angle $45^{\circ}$ and intercept 0.3010 on y – axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gm of adsorbent?

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Solution

The correct option is B 0.6
$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g p$

$\frac{1}{n} = t a n 45^{\circ} l o g k = 0.3010$

$n = 1 k = 2$

$\frac{x}{m} = 2 \times (0.3)^{1}$

$\frac{x}{m} = 0.6$

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When a graph is plotted between log x/m and log P, it is straight line with an angle 45∘ and intercept 0.3010 on y – axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gm of adsorbent?

The correct option is B 0.6logxm=log k+1nlog p 1n=tan 45∘ log k=0.3010 n=1 k=2 xm=2×(0.3)1 xm=0.6

When a graph is plotted between log x/m and log P, it is straight line with an angle $45^{\circ}$ and intercept 0.3010 on y – axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gm of adsorbent?

The correct option is B 0.6
$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g p$

$\frac{1}{n} = t a n 45^{\circ} l o g k = 0.3010$

$n = 1 k = 2$

$\frac{x}{m} = 2 \times (0.3)^{1}$

$\frac{x}{m} = 0.6$