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  • Relative decrease in vapour pressure of an aqueous solution containing 2 moles <img alt="\left [ Cu\left ( NH_{3} \right )_{3} Cl\right ]Cl" src="https://learn.careers360.com/latex-image/?%5Cleft%2

Relative decrease in vapour pressure of an aqueous solution containing 2 moles \left [ Cu\left ( NH_{3} \right )_{3} Cl\right ]Cl in 3 moles H_{2}O\; is 0.5. On reaction with  AgNO_{3} , this solution will form

Option: 1

1\; mol\; AgCl        


Option: 2

0.25\; mol\; AgCl


Option: 3

2\; mol\; AgCl


Option: 4

0.40\; mol\; AgCl


Answers (1)

best_answer

 

Vant Hoff factor (i) -

In case of electrolytes the observed colligative property is different from theoritical colligative property. There ratio is defined by Vant Haff factor

- wherein

i = \frac{observed}{theoritical}

 

Application of Vant Hoff factor -

(a)\: \: \: \left ( \frac{\Delta P}{P^{0}} \right )_{obs}= X_{solute}\times i

(b)\: \: \: \left ( \Delta T_{b} \right )_{obs}= K_{b}\times m\times i

(c)\: \: \: \left ( \Delta T_{f} \right )_{obs}= K_{f}\times m\times i

(d)\: \: \: \left ( \pi \right )_{obs}= C\times R\times T\times i

-

\frac{\Delta P}{P}=\frac{i.n}{i.n+N}\Rightarrow \frac{1}{2}=\frac{i.2}{i.2+3}

So,  i=\frac{3}{2}=1.5\Rightarrow \; 1+\left ( n-1 \right )\alpha

\therefore \; \; \; \; \alpha =\frac{1}{2}

So,          moles of    Cl-\: ions=2\times \frac{1}{2}=1\; moles

So,          moles of  AgCl \; ppt=1\; moles

Posted by

Rakesh

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