The molality of an aqueous solution of glucos | vedclass

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

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$0.0177$

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(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Given $m = 1.0 \ m$,this means $1 \ 	ext{mole}$ of glucose is dissolved in $1 \ 	ext{kg}$ $(1000 \ 	ext{g})$ of water. Moles of glucose $(n_{	ext{glucose}})$ = $1 \ 	ext{mol}$. Moles of water $(n_{	ext{water}})$ = $\frac{1000 \ 	ext{g}}{18 \ 	ext{g/mol}} \approx 55.56 \ 	ext{mol}$. Total moles = $n_{	ext{glucose}} + n_{	ext{water}} = 1 + 55.56 = 56.56 \ 	ext{mol}$. Mole fraction of glucose $(x_{	ext{glucose}})$ = $\frac{n_{	ext{glucose}}}{	ext{Total moles}} = \frac{1}{56.56} \approx 0.0177$.

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

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