Question

The van't hoff factor for 0.1m ba(no3)2 solution is 2.74. The percentage degree of dissociation of ba(no3)2 is

Answer 1

The percentage degree of dissociation of $Ba(NO_3)_2$ is 8.7%

Explanation:

Van't Hoff's factor is the ratio of observed colligative property to the calculated Colligative property.

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

       $Ba(NO_3)_2\rightarrow Ba^{2+}+2NO_3^{-}$

at t= 0             0.1 M                0             0

at eqm  $0.1-\alpha$      $\alpha$       $2\times \alpha$  

Total moles after dissociation=$0.1-\alpha+\alpha+2\times \alpha=0.1+2\times \alpha$  

Thus $i=\frac{0.1+2\times \alpha}{0.1}$

$2.74=\frac{0.1+2\times \alpha}{0.1}$

$\alpha=0.087=0.087\times 100\%=8.7\%$

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Answer 2

correct answer is...

→ 87%

Explanation:

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