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

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(A) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration at the same temperature.Mol

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(A) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration at the same temperature.Molar concentration of urea $= \frac{10 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = \frac{1}{6} \ mol \ L^{-1}$.$A$ $5 \%$ solution of a non-volatile solute means $5 \ g$ of solute in $100 \ mL$ of solution,which is $50 \ g$ of solute in $1 \ L$ $(dm^3)$ of solution.Let the molecular mass of the non-volatile solute be $M$.Molar concentration of the non-volatile solute $= \frac{50 \ g \ L^{-1}}{M \ g \ mol^{-1}} = \frac{50}{M} \ mol \ L^{-1}$.Since the solutions are isotonic,$\frac{1}{6} = \frac{50}{M}$.Therefore,$M = 50 	imes 6 = 300 \ g \ mol^{-1}$.

$A$ solution containing $10 \ g$ 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 this non-volatile solute is ........ $g \ mol^{-1}$.

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