Home
Class 12
CHEMISTRY
In an adsorption experiment, a graph bet...

In an adsorption experiment, a graph between log `(x/m)` versus log `P` was found to be linear with a slope of `45^(@)` . The intercept on the log y axis was found to be `0.301` . Calculate the amount of the gas adsorbed per gram of charcoal under a pressure of `3.0` atm.

A

4

B

2

C

6

D

8

Text Solution

Verified by Experts

The correct Answer is:
C
log `(x)/(m)=logK+(1)/(n)logP`
`:.` Plot of log `(x)/(m)` versus log `P` is linear with slope `=(1)/(n)`
and intercept `=logK`
Thus `(1)/(n)=tantheta=tan45^(@)=1` or `n=1`
log `K=0.301` or `K=` antilog `0.301=2`
At `P=3atm`
`(x)/(m)=KP^(1//n)=2xx(3)^(t)=6`
Promotional Banner

Topper's Solved these Questions

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Fill In The Blanks|37 Videos

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises True /False|20 Videos

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Assertion-Reasoning|17 Videos

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 2.3 (Objective)|9 Videos

  • SYNTHETIC AND NATURAL POLYMERS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Assertion Reasoning|10 Videos

Similar Questions

Explore conceptually related problems

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

In an adsorption experiment a graph between log x/m versus log P was fond to be linear with a slope of 45^@) . The intercept on the log x/m axis was found to 0.70 . Calculate the amount of gas adsorbed per gram of charcoal under a pressure of 1 atm [log5=0.70]

In an Adsorption experiment a graph between log x/m versus log P was found to be linear with a slope of 45^(@) the intercept of the log x/m was found to be 0.3010. Calculate the amount of gas adsorbed per gram of charcoal under a pressure of 0.6 bar.

In an adsorption experiment, a graph of log(x/m) versus log P was found to be linear with a slope of 45°, and the the intercept is found to be 0.3010. The amount of gas adsorbed per gram charcoal under a pressure of 0.8 atm is

Graph between log x/m and log P is a straight line at angle of 45^(@) with intercept 0.4771 on y-axis. Calculate the amount of gas adsorbed in gram per gram of adsorbent when pressure is 3 atm.

When a graph is plotted between log x/m and log p, it is straight line with an angle 45^@ and intercept 0.6020 on y-axis.if initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent :

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 wil be the amount of gas adsorbed per gm of adsorbent :

For the adsorption of a gas on a solid, the plot of log(x/m) versus log P is linear with slope equal to:

For the adsorption of a gas on a solid, the plot of log(x/m) versus log p is linear with slope equal to

Graph between log x//m and log p is a straight line inclined at an angle of 45^@ . When pressure is 0.5 atm and 1n k = 0.693 , the amount of solute adsorbed per gram of adsorbent will be: