The solution containing 18 g dm^{-3} of glu | vedclass

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Question

(D) Since both solutions boil at the same temperature,their elevation in boiling point $(\Delta T_b)$ must be the same.
Since $\Delta T_b = K_b 	ime

Vedclass · Accepted Answer

$60 \ g \ mol^{-1}$

Vedclass · Answer

(D) Since both solutions boil at the same temperature,their elevation in boiling point $(\Delta T_b)$ must be the same.
Since $\Delta T_b = K_b 	imes m$,and $K_b$ is the same for the same solvent (water),the molality $(m)$ of both solutions must be equal.
Given concentration in $g \ dm^{-3}$ is $C = \frac{W}{V}$. For a fixed volume of solvent,the molarity and molality are proportional to the concentration divided by the molar mass.
$\frac{C_{	ext{glucose}}}{M_{	ext{glucose}}} = \frac{C_A}{M_A}$
Substituting the given values:
$\frac{18}{180} = \frac{6}{M_A}$
$0.1 = \frac{6}{M_A}$
$M_A = \frac{6}{0.1} = 60 \ g \ mol^{-1}$.

The solution containing $18 \ g \ dm^{-3}$ of glucose (molar mass $180 \ g \ mol^{-1}$) in water and another containing $6 \ g \ dm^{-3}$ of solute $A$ in water boil at the same temperature. What is the molar mass of $A$?

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