\\\text{Calculate the mole percentage of } {CH}_{3}{OH} \text{ and } \mathrm{H}_{2} \mathrm{O} \text{ respectively in } 60 \% \text{(by mass)} \\ \text{ aqueous solution of } \mathrm{C H}_{3} \mathrm{O H}.

Answers (1)

\\60 \% \text{ by mass solution of } \mathrm{CH}_{3} \mathrm{OH}. \text{ That means in 100 gm solution } \\ 60 \mathrm{gm} \ \mathrm{CH}_{3} \mathrm{OH} \text{ is dissolved in }40 \mathrm{gm} \ \mathrm{H}_{2} \mathrm{O} \\\\ \therefore \text{ Molar mass of } \mathrm{CH}_{3} \mathrm{OH}=32 \mathrm{g} / \mathrm{mol} \\ \therefore \text{Molar mass of }\mathrm{H}_{2} \mathrm{O}=18 \mathrm{g} / \mathrm{mol}

\Rightarrow \text { No. of moles of } \mathrm{CH}_{3} \mathrm{OH}=\frac{60 \mathrm{gm}}{32 \mathrm{g} / \mathrm{mol}}=1.875 \mathrm{mol} . . . . . . . . . .  . .(1)

\text { No. of moles of } \mathrm{H}_{2} \mathrm{O}=\frac{40 \mathrm{gm}}{18 \mathrm{gm} / \mathrm{mol}}=2.222 \mathrm{mol}          . . . . . . . . . . . . (2)

\text { Total no. of moles }=1.875+2.222=4.097 \text { moles } . . . ....... . . . . . . (3)

\Rightarrow \text { Mole percentage }=\frac{\text { Moles of compound }}{\text { Total no. of moles }} \times 100

\begin{array}{l} \text {Mole } \% \text { of } \mathrm{CH}_{3} \mathrm{OH}=\frac{1.875}{4.097} \times 100 \\ =45.8 \% \quad \text { (From } 1 \& 3) \end{array}

\begin{array}{l} \text { Mole } \% \text { of } \mathrm{H}_{2} \mathrm{O}=\frac{2.222}{4.097} \times 100 \\ =54.23 \% \end{array}

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