A 25 % solution of cane-sugar (molar mass = 3 | vedclass

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(B) Two solutions are isotonic when they have the same osmotic pressure,which implies they have the same molar concentration.For a $25 \%$ solution of

Vedclass · Accepted Answer

$68.4 \ g \ mol^{-1}$

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(B) Two solutions are isotonic when they have the same osmotic pressure,which implies they have the same molar concentration.For a $25 \%$ solution of cane-sugar,$100 \ g$ of solution contains $25 \ g$ of cane-sugar.Assuming the density of the solution is $1 \ g \ mL^{-1}$,$100 \ g$ of solution is approximately $100 \ mL$.Molarity of cane-sugar $= \frac{25 \ g}{342 \ g \ mol^{-1}} 	imes \frac{1000 \ mL}{100 \ mL} = \frac{250}{342} \ M \approx 0.7309 \ M$.For a $5 \%$ solution of substance $A$,$100 \ g$ of solution contains $5 \ g$ of substance $A$.Let the molar mass of $A$ be $M_A$.Molarity of $A = \frac{5 \ g}{M_A \ g \ mol^{-1}} 	imes \frac{1000 \ mL}{100 \ mL} = \frac{50}{M_A} \ M$.Since the solutions are isotonic,their molarities are equal:$\frac{50}{M_A} = \frac{250}{342}$.$M_A = \frac{50 	imes 342}{250} = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.

$A$ $25 \%$ solution of cane-sugar (molar mass $= 342 \ g \ mol^{-1}$) is isotonic with a $5 \%$ solution of a substance $A$. Find the molecular weight of $A$.

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