Home
Class 12
CHEMISTRY
We have three aqueous solutions of NaCl ...

We have three aqueous solutions of NaCl labelled as A, B and C with concentration `0.1 M , 0.01 "and" 0.001` M , respectively . The value of van't Hoff factor for these solutions will be in the order :

A

`i_(A)lti_(B)lti_(C)`

B

`i_(A)gti_(B)gti_(C)`

C

`i_(A)=i_(B)=i_(C)`

D

`i_(A)lti_(B)gti_(C)`

Text Solution

Verified by Experts

The correct Answer is:
A

Van’t Hoff’s factor i depends on the degree of dissociation `(alpha)` and the value of `alpha` increases with the increase of dilution of an electrolyte solution
So `alpha_(A)ltalpha_(B)ltalpha_(C)`
or `i_(A)lti_(B)lti_(C)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

The Van't Hoff factor for a dillute aqueous solution of glucose is

At 298 K, the ratio of osmotic pressure of two solutions of a substance with concentrations of 0.01 M and 0.001 M, respectively, is

Four solutions of K_(2)SO_(4) with the following concentration 0.1 m, 0.01 m, 0.001 m and 0.0001 m are available. The maximum value of Van’t Hoff factor (i) will be of: