0.004 M Na_2SO_4 solution is isotonic with 0 | vedclass

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(C) Since the solutions are isotonic,their osmotic pressures are equal: $\pi_{Na_2SO_4} = \pi_{Glucose}$.For glucose (non-electrolyte),$\pi = C 	imes

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) Since the solutions are isotonic,their osmotic pressures are equal: $\pi_{Na_2SO_4} = \pi_{Glucose}$.For glucose (non-electrolyte),$\pi = C 	imes R 	imes T = 0.01 	imes R 	imes T$.For $Na_2SO_4$,$\pi = i 	imes C 	imes R 	imes T = i 	imes 0.004 	imes R 	imes T$.Equating the two: $i 	imes 0.004 = 0.01$,so $i = 2.5$.For $Na_2SO_4 ightarrow 2 Na^{+} + SO_4^{2-}$,the van't Hoff factor $i = 1 + (n-1)\alpha$,where $n = 3$.$2.5 = 1 + (3-1)\alpha = 1 + 2\alpha$.$2\alpha = 1.5$,which gives $\alpha = 0.75$.Therefore,the degree of dissociation is $75 \%$.

$0.004 \ M$ $Na_2SO_4$ solution is isotonic with $0.01 \ M$ glucose solution,then the degree of dissociation of $Na_2SO_4$ is ........... $\%$.

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