The molar mass of a solute X in g mol^{-1}, | vedclass

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(A) For isotonic solutions, the osmotic pressure is equal, so the molar concentrations are equal: $\frac{W_1}{M_1 V_1} = \frac{W_2}{M_2 V_2}$.Given th

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$68.4$

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(A) For isotonic solutions, the osmotic pressure is equal, so the molar concentrations are equal: $\frac{W_1}{M_1 V_1} = \frac{W_2}{M_2 V_2}$.Given that the solutions are $1 \%$ and $5 \%$ by mass, we can assume $1 \ g$ of solute $X$ in $100 \ mL$ solution and $5 \ g$ of cane sugar in $100 \ mL$ solution.Here, $W_1 = 1 \ g$, $W_2 = 5 \ g$, $M_2 = 342 \ g \ mol^{-1}$, and $V_1 = V_2 = 100 \ mL$.Substituting the values: $\frac{1}{M_1 	imes 100} = \frac{5}{342 	imes 100}$.$M_1 = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.

The molar mass of a solute $X$ in $g \ mol^{-1}$, if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $= 342 \ g \ mol^{-1}$), is

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