A 5% (w/v) solution of anhydrous CaCl_2 at 0 | vedclass

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(C) Given: $w = 5 \ g$,$V = 100 \ mL = 0.1 \ L$,$T = 273 \ K$,$\pi = 15 \ atm$,$M_{CaCl_2} = 111 \ g/mol$. Using the formula $\pi = iCRT$,where $C = \

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$24.33$

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(C) Given: $w = 5 \ g$,$V = 100 \ mL = 0.1 \ L$,$T = 273 \ K$,$\pi = 15 \ atm$,$M_{CaCl_2} = 111 \ g/mol$. Using the formula $\pi = iCRT$,where $C = \frac{n}{V} = \frac{w}{M 	imes V} = \frac{5}{111 	imes 0.1} = 0.4504 \ M$. $15 = i 	imes 0.4504 	imes 0.0821 	imes 273$. $15 = i 	imes 10.09$. $i = \frac{15}{10.09} = 1.4866$. For $CaCl_2 ightarrow Ca^{2+} + 2Cl^-$,$n = 3$. Degree of dissociation $\alpha = \frac{i - 1}{n - 1} = \frac{1.4866 - 1}{3 - 1} = \frac{0.4866}{2} = 0.2433$. Therefore,$\alpha = 24.33\%$.

$A$ $5\% \ (w/v)$ solution of anhydrous $CaCl_2$ at $0^o C$ exerts an osmotic pressure of $15 \ atm$. The degree of dissociation $(\alpha)$ of $CaCl_2$ is ....... $\%$.

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