A solution containing 10 g per dm^3 of urea | vedclass

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(C) For isotonic solutions,the osmotic pressures are equal,so $\pi_1 = \pi_2$. Since the temperature is constant,the molar concentrations are equal: $

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$300$

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(C) For isotonic solutions,the osmotic pressures are equal,so $\pi_1 = \pi_2$. Since the temperature is constant,the molar concentrations are equal: $C_1 = C_2$. For urea: Concentration $C_1 = \frac{10 \ g}{60 \ g \ mol^{-1} 	imes 1 \ L} = \frac{1}{6} \ mol \ L^{-1}$. For the non-volatile solute: $A$ $5 \%$ solution means $5 \ g$ of solute in $100 \ mL$ of solution,which is $50 \ g \ L^{-1}$. So,$C_2 = \frac{50}{M_w} \ mol \ L^{-1}$,where $M_w$ is the molecular mass. Equating the concentrations: $\frac{1}{6} = \frac{50}{M_w}$. Solving for $M_w$: $M_w = 50 	imes 6 = 300 \ g \ mol^{-1}$.

$A$ solution containing $10 \ g$ per $dm^3$ of urea (molecular mass $= 60 \ g \ mol^{-1}$) is isotonic with a $5 \%$ solution of a non-volatile solute. The molecular mass of the non-volatile solute is ........ $g \ mol^{-1}$.

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