In an adsorption experiment- a graph between log x - m versus log p is found to be linear with a slope of 45- The intercept on log x - m axis was found to be 0-3010- The amount of the gas adsorbed per gram of charcoal under a pressure of 0-5 atm will beA- 1B- 2C- 0-5D- 0-25

The correct option is A 1log x/m=log k+1/nlog p 1/n=tan θ=tan 45^∘=1 log k=0.301⇒ k=2 At p=0.5 atm x/m=K P^1/n=2(0.5)^1 =1 ...

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Standard XII

Chemistry

Introduction

In an adsorpt...

Question

In an adsorption experiment, a graph between log(x/m) versus log p is found to be linear with a slope of $45^{\circ}$ . The intercept on log (x/m) axis was found to be 0.3010. The amount of the gas adsorbed per gram of charcoal under a pressure of 0.5 atm will be

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Solution

The correct option is A 1
$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g p$
$\frac{1}{n} = t a n θ = t a n 45^{\circ} = 1$
$l o g k = 0.301 \Rightarrow k = 2$
$A t p = 0.5 a t m$
$\frac{x}{m} = K P^{\frac{1}{n}} = 2 (0.5)^{1}$
$= 1$

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In an adsorption experiment, a graph between log(x/m) versus log p is found to be linear with a slope of 45∘. The intercept on log (x/m) axis was found to be 0.3010. The amount of the gas adsorbed per gram of charcoal under a pressure of 0.5 atm will be

The correct option is A 1log xm=log k+1nlog p 1n=tan θ=tan 45∘=1 log k=0.301⇒k=2 At p=0.5 atm xm=K P1n=2(0.5)1 =1

In an adsorption experiment, a graph between log(x/m) versus log p is found to be linear with a slope of $45^{\circ}$ . The intercept on log (x/m) axis was found to be 0.3010. The amount of the gas adsorbed per gram of charcoal under a pressure of 0.5 atm will be

The correct option is A 1
$l o g \frac{x}{m} = l o g k + \frac{1}{n} l o g p$
$\frac{1}{n} = t a n θ = t a n 45^{\circ} = 1$
$l o g k = 0.301 \Rightarrow k = 2$
$A t p = 0.5 a t m$
$\frac{x}{m} = K P^{\frac{1}{n}} = 2 (0.5)^{1}$
$= 1$