If K_{b} denotes the molal elevation constant | vedclass

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Question

(A) The molality $(m)$ of the solution is calculated as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.Given mass of glucose $= 36

Vedclass · Accepted Answer

$100 + 0.2 K_{b}$

Vedclass · Answer

(A) The molality $(m)$ of the solution is calculated as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.Given mass of glucose $= 36 \ g$,molar mass $= 180 \ g/mol$.Moles of glucose $= \frac{36}{180} = 0.2 \ mol$.Assuming the density of the solution is approximately $1 \ g/mL$,$1 \ dm^3$ (or $1 \ L$) of solution contains $1000 \ g$ of water,which is $1 \ kg$.Thus,$m = 0.2 \ mol / 1 \ kg = 0.2 \ m$.The elevation in boiling point is given by $\Delta T_{b} = K_{b} 	imes m = K_{b} 	imes 0.2 = 0.2 K_{b}$.The boiling point of the solution $= 100 + \Delta T_{b} = 100 + 0.2 K_{b}$.

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 $= 180 \ g/mol$) per $1 \ kg$ of water is:

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