A solution containing 10 g of urea (molar ma | vedclass

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(B) For isotonic solutions,the osmotic pressure $(\pi)$ is equal,which means their molar concentrations $(C)$ are equal at the same temperature.$C_1 =

Vedclass · Accepted Answer

$300$

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(B) For isotonic solutions,the osmotic pressure $(\pi)$ is equal,which means their molar concentrations $(C)$ are equal at the same temperature.$C_1 = C_2$$\frac{n_1}{V_1} = \frac{n_2}{V_2}$Given for urea: Mass $= 10 \ g$,Molar mass $= 60 \ g \ mol^{-1}$,Volume $= 1 \ L$.$n_1 = \frac{10}{60} = \frac{1}{6} \ mol$.$C_1 = \frac{1/6}{1} = \frac{1}{6} \ mol \ L^{-1}$.Given for non-volatile solute: $5 \% \ w/v$ means $5 \ g$ in $100 \ mL$,which is $50 \ g$ in $1 \ L$.Let the molar mass be $M_2$.$n_2 = \frac{50}{M_2}$.$C_2 = \frac{50}{M_2} \ mol \ L^{-1}$.Equating $C_1 = C_2$:$\frac{1}{6} = \frac{50}{M_2}$$M_2 = 50 	imes 6 = 300 \ g \ mol^{-1}$.

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

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