Home
Class 12
CHEMISTRY
The van't Hoff factor for 0.1 M Ba(NO(3)...

The van't Hoff factor for 0.1 M `Ba(NO_(3))_(2)` solution is 2.74 The degree of dissociation is :

A

`91.3%`

B

`87%`

C

`100%`

D

`74%`

Text Solution

Verified by Experts

The correct Answer is:
B
`Ba(NO_(3))_(2) hArr Ba^(2+)+2NO^(3-)`
`{:(1,0,0),((1-alpha), alpha,alpha):}`
`i=1-alpha+alpha+2alpha=(1+2alpha)`
`i=(1+2alpha)`
`:.alpha=((i-1))/(2)=(2.74-1)/(2)=0.87`
`=87%`
Hence, (B) is the correct answer.
Promotional Banner

Similar Questions

Explore conceptually related problems

The Van't Hoff factor of a 0.1 M Al_(2)(SO_(4))_(3) solution is 4.20 . The degree of dissociation is

The Van't Hoff factor 0.1M Ba(NO_(3))_(2) solution is found to be 2.74 the percentage dissociation of the salt is:

Van't Hoff factor of Ca(NO_(3))_(2) is

The van't Hoff factor of a 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is :

Discuss the van't Hoff factor.

If van't Hoff factor i=1,then

The van't Hoff factor of BaCl_2 at 0.01 M concentration is 1.98. The percentage of dissociation of BaCl_2 at this concentration is:

Van't Hoff factor of 0.1 m aqueous solution of Al_2(SO_4)_3 which undergoes 80% dissociation is

Calculate the Van't Hoff factor when 0.1 mol NH_(4)Cl is dissolved in 1 L of water. The degree of dissociation of NH_(4)Cl is 0.8 and its degree of hydrolysis is 0.1 .

The van't Hoff factor (i) accounts for