(C) For isotonic solutions,the osmotic pressures are equal,which implies their molar concentrations are equal: $C_1 = C_2$. The molar concentration $C

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(C) For isotonic solutions,the osmotic pressures are equal,which implies their molar concentrations are equal: $C_1 = C_2$. The molar concentration $C$ is given by $\frac{\text{mass concentration (g/L)}}{\text{molar mass (g/mol)}}$. For cane sugar: $C_1 = \frac{50 \text{ g/L}}{342 \text{ g/mol}}$. For the unknown solute: $C_2 = \frac{10 \text{ g/L}}{M_2}$. Equating the two: $\frac{50}{342} = \frac{10}{M_2}$. Solving for $M_2$: $M_2 = \frac{10 \times 342}{50} = \frac{342}{5} = 68.4 \text{ g/mol}$.

Class 12 Chemistry Solutions Osmosis and Osmotic pressure of the solution

Medium

$A$ $5\% \text{ (w/v)}$ solution of cane sugar (molar mass = $342 \text{ g/mol}$) is isotonic with a $1\% \text{ (w/v)}$ solution of an unknown solute. What is the molar mass of the unknown solute in $\text{g/mol}$?

A
$136.2$
B
$171.2$
C
$68.4$
D
$34.2$

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Osmosis and Osmotic pressure of the solution

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

EasyAP EAMCET 2014

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If the osmotic pressure of cane sugar solution is $2.46 \ atm$ at $27^{\circ} C$,then what is the concentration (in $mol \ L^{-1}$) of the solution $(R=0.0821 \ L \ atm \ mol^{-1} \ K^{-1})$?

MediumAP EAMCET 2023

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If a $6.84 \% (wt./vol.)$ solution of cane sugar $(mol. wt. 342)$ is isotonic with a $1.52 \% (wt./vol.)$ solution of thiocarbamide,then the molecular weight of thiocarbamide is:

Medium

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$A$ $0.002 \, M$ solution of $NaCl$ having a degree of dissociation of $90 \%$ at $27 \, ^{\circ}C$ has an osmotic pressure equal to:

Medium

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$A$ 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 ........ $g \ mol^{-1}$.

MediumAIIMS 2014

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$A$ $5\% \text{ (w/v)}$ solution of cane sugar (molar mass = $342 \text{ g/mol}$) is isotonic with a $1\% \text{ (w/v)}$ solution of an unknown solute. What is the molar mass of the unknown solute in $\text{g/mol}$?

Similar Questions

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

If the osmotic pressure of cane sugar solution is $2.46 \ atm$ at $27^{\circ} C$,then what is the concentration (in $mol \ L^{-1}$) of the solution $(R=0.0821 \ L \ atm \ mol^{-1} \ K^{-1})$?

If a $6.84 \% (wt./vol.)$ solution of cane sugar $(mol. wt. 342)$ is isotonic with a $1.52 \% (wt./vol.)$ solution of thiocarbamide,then the molecular weight of thiocarbamide is:

$A$ $0.002 \, M$ solution of $NaCl$ having a degree of dissociation of $90 \%$ at $27 \, ^{\circ}C$ has an osmotic pressure equal to:

$A$ 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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