A solution containing 10 , g , dm^{-3} of ure | vedclass

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Question

(B) Two solutions are isotonic if they have the same molar concentration $(C_1 = C_2)$.For urea: Concentration $= \frac{10 \, g \, dm^{-3}}{60 \, g \,

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) Two solutions are isotonic if they have the same molar concentration $(C_1 = C_2)$.For urea: Concentration $= \frac{10 \, g \, dm^{-3}}{60 \, g \, mol^{-1}} = \frac{1}{6} \, mol \, dm^{-3}$.For the non-volatile solute: $A$ $5 \%$ solution means $5 \, g$ of solute in $100 \, mL$ of solution,which is equivalent to $50 \, g \, dm^{-3}$.Let the molecular mass of the non-volatile solute be $M$.Concentration of non-volatile solute $= \frac{50 \, g \, dm^{-3}}{M \, g \, mol^{-1}} = \frac{50}{M} \, mol \, dm^{-3}$.Since the solutions are isotonic,$\frac{1}{6} = \frac{50}{M}$.Therefore,$M = 50 	imes 6 = 300 \, g \, mol^{-1}$.

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

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