If a solution of methyl alcohol (CH_3OH) is 5 | vedclass

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(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Given $m = 5.2 \ mol/kg$,this means there are $5.2 \ moles$ of

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$0.08$

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(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Given $m = 5.2 \ mol/kg$,this means there are $5.2 \ moles$ of $CH_3OH$ in $1000 \ g$ $(1 \ kg)$ of water. Moles of water $(n_{H_2O})$ = $\frac{1000 \ g}{18.02 \ g/mol} \approx 55.55 \ moles$. Mole fraction of $CH_3OH$ $(x_{CH_3OH})$ = $\frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}}$. $x_{CH_3OH} = \frac{5.2}{5.2 + 55.55} = \frac{5.2}{60.75} \approx 0.0856$. Rounding to the nearest provided option,the value is approximately $0.08$.

If a solution of methyl alcohol $(CH_3OH)$ is $5.2 \ m$ (molal) in water,what is the mole fraction of methyl alcohol in the solution?

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