A 5% (w/v) solution of anhydrous ${CaCl}_{2} at 0^{o} c$ developed 15 atm. osmotic pressure . What is the degree of dissociation of ${CaCl}_{2}$ .

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$24.3 %$

$2.43 %$

$52.4 %$

$5.24 %$

answer is D.

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Detailed Solution

$π = CST . i 15 = \frac{5}{111} \times \frac{1000}{100} \times 0.0821 \times 273 \times i i = \frac{15 \times 111}{5 \times 10 \times 0.0821 \times 273} i = 1.48$

$α = \frac{i - 1}{n - 1} α = \frac{1.48 - 1}{3 - 1} = \frac{0.48}{2} = 0.24 = 24 %$

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