The molality of an aqueous solution of glucos | vedclass

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Question

(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given $m = 1.0 \ m$,this means $1 \ mol$ of glucose is dissolve

Vedclass · Accepted Answer

$18/1000$

Vedclass · Answer

(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given $m = 1.0 \ m$,this means $1 \ mol$ of glucose is dissolved in $1 \ kg$ $(1000 \ g)$ of water.Moles of glucose $(n_2)$ = $1 \ mol$.Moles of water $(n_1)$ = $\frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.Mole fraction of glucose $(X_2)$ = $\frac{n_2}{n_1 + n_2} = \frac{1}{55.56 + 1} = \frac{1}{56.56} \approx 0.0177$.Approximating $56.56 \approx 55.56$,we get $X_2 \approx \frac{1}{55.56} \approx 0.018 = \frac{18}{1000}$.

The molality of an aqueous solution of glucose is $1.0 \ m$. The mole fraction of glucose is approximately ........

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