Question

The vant hoff factor of bacl2 at 0.01m concentration is 1.98.The percentage of dissociation of bacl2 at this concentration is

Answer 1

Answer: 49%

Explanation:

Vant hoff factor is the ratio of observed colligative property to the calculated colligative property.

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

$BaCl_2\rightarrow Ba^{2+}++2Cl^{-}$

0.01- M            0             0

$0.01-0.01\alpha$       $0.01\alpha$        $0.02\alpha$  

Total moles after dissociation =$0.01-0.01\alpha+0.01\alpha+0.02\alpha=0.01+2\alpha$  

thus $i=\frac{0.01+0.02\alpha}{0.01}$

$1.98=\frac{0.01+0.02\alpha}{0.01}$

$\alpha=0.49$

Thus percent dissociation is 49%

Answer 2

Answer:49%

Explanation:

Hey Guys,  

Given Data: Molarity ofBaCl2=0.01M

  and i =1.98

Solution Down below,

 $\alpha$(alpha)=i-1/n-1

BaCl2Ba+2+2cl-1

 n=3-1=2

$\alpha$=(1.98-1)/(2-1)=0.49

In percentage 0.49x100=49%