You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Types of Catalysis

The graph bet...

Question

The graph between log $\frac{X}{m}$ v/s log P is straight-line inclined at an angle $45^{o}$ with intercept 0.30. What will be the rate of adsorption at pressure of 0.4 atm?

0.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.8
Slope in equation is $\frac{1}{n}$ = tan $45^{o}$ =1
Intercept is logk = 0.30, that implies k=2
Hence by equation $\frac{x}{m} = k p^{1 / n}$
k=2, n=1 and p=0.4

$∴$ Rate of adsorption, x/m is 0.8.
Option C is correct.

Suggest Corrections

Similar questions

Q. The graph between

log \frac{x}{m} v / s log P

is straight line inclined at an angle

45^{\circ}

with intercept

0.30

. What will be rate of adsorption at pressure

0.4 a t m

Q. Graph between log

\frac{x}{m}

and log

P

is a straight line inclined at an angle

45^{o}

. When pressure of

0.5 a t m

and

l o g k = 0.699,

the amount of solute adsorbed per

g

of adsorbent will be (Given, antilog 0.699 = 5)

Q. In an adsorption experiment, a graph between log

(\frac{x}{m})

versus log

p

is found to be linear with slope of

45^{o}

. The intercept on log

(\frac{x}{m})

axis was found to

0.3010

. The amount of the gas adsorbed per gram of charcoal under the pressure of

0.5

atm will be:

Q. Graph between log

(\frac{x}{m})

and

l o g p

is a straight line at an angle

45^{\circ}

with intercept on y-axis at 0.3010. Calculate the amount of gas adsorbed in gram per gram of the adsorbent when pressure is

0.2 a t m

Q. Graph between

log (\frac{x}{m})

and

log P

is a straight line at an angle

45^{\circ}

with intercept on

y

-axis

0.3010

. The amount (in

g

) of the gas absorbed per

g

of the adsorbent when pressure is

0.2 a t m

is:
(Assume that the adsorption obey Freundlich isotherm)

Join BYJU'S Learning Program

The graph between log Xm v/s log P is straight-line inclined at an angle 45o with intercept 0.30. What will be the rate of adsorption at pressure of 0.4 atm?

The correct option is C 0.8 Slope in equation is 1n = tan45o =1 Intercept is logk = 0.30, that implies k=2 Hence by equation xm=kp1/n k=2, n=1 and p=0.4∴ Rate of adsorption, x/m is 0.8.Option C is correct.

The graph between log $\frac{X}{m}$ v/s log P is straight-line inclined at an angle $45^{o}$ with intercept 0.30. What will be the rate of adsorption at pressure of 0.4 atm?

The correct option is C 0.8
Slope in equation is $\frac{1}{n}$ = tan $45^{o}$ =1
Intercept is logk = 0.30, that implies k=2
Hence by equation $\frac{x}{m} = k p^{1 / n}$
k=2, n=1 and p=0.4

$∴$ Rate of adsorption, x/m is 0.8.
Option C is correct.