If 0.004 M Na_2SO_4 and 0.01 M glucose aqu | vedclass

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(C) For isotonic solutions,the osmotic pressure is equal,so $i_1 C_1 = i_2 C_2$. For $Na_2SO_4$,the van't Hoff factor $i = 1 + (n - 1)\alpha$,where $n

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$75$

Vedclass · Answer

(C) For isotonic solutions,the osmotic pressure is equal,so $i_1 C_1 = i_2 C_2$. For $Na_2SO_4$,the van't Hoff factor $i = 1 + (n - 1)\alpha$,where $n = 3$ $(Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-})$. For glucose,$i = 1$ as it is a non-electrolyte. Substituting the values: $[1 + (3 - 1)\alpha] 	imes 0.004 = 1 	imes 0.01$. $[1 + 2\alpha] = \frac{0.01}{0.004} = 2.5$. $2\alpha = 2.5 - 1 = 1.5$. $\alpha = \frac{1.5}{2} = 0.75$. Therefore,the degree of dissociation in percentage is $0.75 	imes 100 = 75 \%$.

If $0.004 \ M \ Na_2SO_4$ and $0.01 \ M$ glucose aqueous solutions are isotonic,then the degree of dissociation of $Na_2SO_4$ is ........ $\%$.

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