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

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(A) For isotonic solutions,the osmotic pressures are equal,so $\frac{W_1}{V_1 M_1} = \frac{W_2}{V_2 M_2}$.Given that the solutions are $1 \%$ and $5 \

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$68.4$

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(A) For isotonic solutions,the osmotic pressures are equal,so $\frac{W_1}{V_1 M_1} = \frac{W_2}{V_2 M_2}$.Given that the solutions are $1 \%$ and $5 \%$,we have $\frac{W_1}{V_1} = 1 \ g/100 \ mL$ and $\frac{W_2}{V_2} = 5 \ g/100 \ mL$.Let $M_1$ be the molar mass of solute $X$ and $M_2 = 342 \ g \ mol^{-1}$ be the molar mass of cane sugar.Substituting the values: $\frac{1}{M_1} = \frac{5}{342}$.Solving for $M_1$: $M_1 = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.

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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