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Question

(D) The molality $m$ of the solution is calculated as follows:Mass of glucose $(W_2)$ $= 36 \ g$.Molar mass of glucose $(M_2)$ $= 180 \ g \ mol^{-1}$.

Vedclass · Accepted Answer

$(100 + \frac{2 \ K_b}{10})^{\circ} C$

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(D) The molality $m$ of the solution is calculated as follows:Mass of glucose $(W_2)$ $= 36 \ g$.Molar mass of glucose $(M_2)$ $= 180 \ g \ mol^{-1}$.Volume of solution $= 1 \ dm^3 = 1 \ L$.Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the mass of the solvent (water) $W_1$ is $1000 \ g = 1 \ kg$.Molality $m = \frac{W_2}{M_2 	imes W_1 (	ext{in } kg)} = \frac{36}{180 	imes 1} = 0.2 \ mol \ kg^{-1} = \frac{2}{10} \ mol \ kg^{-1}$.The elevation in boiling point is given by $\Delta T_b = K_b 	imes m = K_b 	imes \frac{2}{10} = \frac{2 \ K_b}{10}$.The boiling point of water is $100^{\circ} C$.Therefore,the boiling point of the solution $T_b = 100 + \Delta T_b = (100 + \frac{2 \ K_b}{10})^{\circ} C$.

If $K_b$ denotes the molal elevation constant of water,then the boiling point of an aqueous solution containing $36 \ g$ of glucose (molar mass $= 180 \ g \ mol^{-1}$) per $dm^3$ is:

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