A 1% (wt/vol) KCl solution is ionised to th | vedclass

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(B) The osmotic pressure is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constan

Vedclass · Accepted Answer

$5.94$

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(B) The osmotic pressure is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.For $KCl$,the dissociation is $KCl ightarrow K^{+} + Cl^{-}$,so $n = 2$.Given the degree of dissociation $\alpha = 0.8$,the van't Hoff factor is $i = 1 + \alpha(n - 1) = 1 + 0.8(2 - 1) = 1.8$.$A$ $1\% \ (wt/vol)$ solution means $1 \ g$ of $KCl$ in $100 \ mL$ of solution,which is $10 \ g/L$.The molar mass of $KCl$ is $39.1 + 35.5 = 74.6 \ g/mol$.Concentration $C = \frac{10 \ g/L}{74.6 \ g/mol} \approx 0.134 \ mol/L$.Temperature $T = 27 + 273 = 300 \ K$.$\pi = 1.8 	imes 0.134 	imes 0.0821 	imes 300 \approx 5.94 \ atm$.

$A$ $1\% \ (wt/vol) \ KCl$ solution is ionised to the extent of $80\%$. The osmotic pressure at $27 \ ^oC$ of the solution will be .......... $atm$.

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