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A solution containing $1.9g\: \: per\: \: 100mL\: \: of\: \: KCl\left ( M=74.5gmol^{-1} \right )$ is isotonic with a solution containing $3g\: per\: 100mL\: \: of\: \: urea\left ( M=60gmol^{-1} \right )$ . Calculate the degree of dissociation of KCl solution. Assume that both the solutions have same temperature.

Answers (1)

For isotonic solutions,

$C_{KCl}RT=C_{urea}$

$\Rightarrow i\times C_{KCl}=C_{urea}$

$\Rightarrow i\times\frac{1.9\times 1000}{74.5\times 100}=\frac{3}{60}\times\frac{1000}{100}$

$\Rightarrow i\times 0.25=0.5$

$\Rightarrow i=2$

$i=\frac{observed\: colligative\:property}{calculated\:colligative\:property}$

$KCl\rightarrow \: \: \: \: \: K^+\: \: +\: \: \: Cl^-$

$0.25$ $0$ $0$

$0.25-\alpha$ $\alpha$ $\alpha$

Total moles after dissociation = $0.25-\alpha+\alpha +\alpha$

$=0.25+\alpha$

$i=\frac{0.25+\alpha}{0.25}\Rightarrow 2=\frac{0.25+\alpha}{0.25}\Rightarrow \alpha=0.25$

Posted by

Sumit Saini

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