A solution of 12 g of non-electrolyte (A) pr | vedclass

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(C) For two solutions at the same temperature,if osmotic pressure $\pi_1 = \pi_2$,then their molar concentrations must be equal $(C_1 = C_2)$.Given,$C

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(C) For two solutions at the same temperature,if osmotic pressure $\pi_1 = \pi_2$,then their molar concentrations must be equal $(C_1 = C_2)$.Given,$C_2 = 0.05 \ M$ (glucose).For solution $A$,mass $w = 12 \ g$ and volume $V = 1000 \ mL = 1 \ L$.Concentration $C_1 = \frac{n}{V} = \frac{w / M_A}{1 \ L} = \frac{12}{M_A} \ M$.Equating the concentrations: $\frac{12}{M_A} = 0.05$.$M_A = \frac{12}{0.05} = 240 \ g \ mol^{-1}$.Thus,the molecular mass of $A$ is $240 \ g \ mol^{-1}$.

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