A 5.2 molal aqueous solution of methyl alcoho | vedclass

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(C) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given $m = 5.2 \ mol/kg$,this means $5.2$ moles of $CH_3OH$ are

Vedclass · Accepted Answer

$0.086$

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(C) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Given $m = 5.2 \ mol/kg$,this means $5.2$ moles of $CH_3OH$ are dissolved in $1000 \ g$ of water $(H_2O)$.Moles of water $(n_{H_2O})$ $= \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.Mole fraction of methyl alcohol $(X_{CH_3OH})$ $= \frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}}$.$X_{CH_3OH} = \frac{5.2}{5.2 + 55.56} = \frac{5.2}{60.76} \approx 0.0856$.Rounding to two significant figures,we get $0.086$.

$A$ $5.2$ molal aqueous solution of methyl alcohol,$CH_3OH$,is supplied. What is the mole fraction of methyl alcohol in the solution?

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