Question

a solution of substance containin 1.05g per 100 ml was found to be isotonic with 3% W/V glucose solution.the molecular mass of substance is ?
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Answer 1

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Given that the unknown solution

is isotonic with glucose solution

That means there osmotic

pressure is same.

π=CRT=M2VW2×RT

(M2VW2×RT)unknown=(M2VW2×RT)glucose

(M2×0.11.05×RT)unknown=(180×0.13×RT)glucose

M2=0.1×3180×0.1×1.05=63

molecular mass of the unknown solute =63g