Question

A solution contains $25$ and 50% $C{H}_{3}COOH$ by mass. The mole fraction of $H_{2}O$ would be

• A. 0.25
• B. 2.5
• C. 0.502
• D. 5.03

Answer 1

Answer: (C)

Mole fraction of $H_{2}O$ $=\frac{numberofmolesof}{totalnumberofmolesofallcomponents}$ Let the total mass of solution $=100g$ Mass of $H_{2}O=25g$ Mass of $C_2{H}_{5}OH=25g$ Mass of $C{H}_{3}COOH=50g$ Moles of $H_{2}O=\frac{25}{18}=1.388$ ( $\because$ Molar mass of $H_{2}O=18$ ) Moles of $C_2{H}_{5}OH=\frac{25}{46}=0.543$ ( $\because$ Molar mass $C_2{H}_{5}OH=46$ ) Moles of $C{H}_{3}COOH=\frac{50}{60}=0.833$ ( $\because$ Molar mass of $C{H}_{3}COOH=60$ ) Total no. of moles $=1.388+0.543+0.833$ $=2.764$ $\therefore$ Mole fraction of $H_{2}O=\frac{1.388}{2.764}$ $=0.502$