JEE MainChemistryModeratePhysical Chemistry

A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution
of glucose at same temperature. The apparent percentage dissociation of
Na2SO4 is

Show answer and solution

\left( \pi_{obs\ } \right)_{{Na}_{2}{SO}_{4}} = \pi_{glucose\ }

\frac{10}{4} = \frac{1 + 2\alpha}{1}

10 = 4 + 8\alpha

\alpha = \frac{10 - 4}{8} = 0.75

\%\ of\ \alpha = 75\%

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx

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