If 0.4M $A l_{2} (S O_{4})_{3}$ solution at 300 K is found to isotonic with 0.5 M NaCl ( $100 %$ dissociation) solution, calculate degree of dissociation of $A l_{2} (S O_{4})_{3}$ .

50 %

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25 %

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17.2 %

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37.5 %

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Solution

The correct option is D $37.5 %$
$(i C R T)_{A l_{2} (S O_{4})_{2}} = (i C R T)_{N a C l}$
$i \times 0.4 = 2 \times 0.5$
$i = \frac{5}{2}$
Also,
$A l_{2} (S O_{4} 1 1 - α)_{3} \to 2 A l^{3 +} 0 2 α + 3 S O_{4}^{2 -} 0 3 - α$
$i = 1 + 4 α$
$\Rightarrow α = \frac{3}{8}$
and $%$ of dissociation $= \frac{3}{8} \times 100 = 37.5 %$

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If 0.4M Al2(SO4)3 solution at 300 K is found to isotonic with 0.5 M NaCl (100% dissociation) solution, calculate degree of dissociation of Al2(SO4)3.

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