Osmotic pressure of a 4.44% solution of anhyd | vedclass

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(B) Given: $\pi = 16.42 \ atm$,$T = 300 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.Concentration $C = \frac{4.44 \ g}{111 \ g/mol} 	imes \frac{10

Vedclass · Accepted Answer

$33$

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(B) Given: $\pi = 16.42 \ atm$,$T = 300 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.Concentration $C = \frac{4.44 \ g}{111 \ g/mol} 	imes \frac{1000 \ mL}{100 \ mL} = 0.4 \ M$.Using the formula $\pi = iCRT$,we find the van't Hoff factor $i = \frac{\pi}{CRT} = \frac{16.42}{0.4 	imes 0.082 	imes 300} = \frac{16.42}{9.84} \approx 1.668 \approx 1.67$.For $CaCl_2 \longrightarrow Ca^{2+} + 2Cl^-$,the number of ions $n = 3$.The degree of dissociation $\alpha$ is given by $i = 1 + (n-1)\alpha$.$1.67 = 1 + (3-1)\alpha \implies 0.67 = 2\alpha \implies \alpha = 0.335$.Thus,the degree of dissociation is approximately $33.5 \%$,which corresponds to $33 \%$.

Osmotic pressure of a $4.44\%$ solution of anhydrous $CaCl_2$ was found to be $16.42 \ atm$ at $27 \, ^\circ C$. What is the degree of dissociation of $CaCl_2$ in $\%$?

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