Question

A mixture has 18 g water and 414 g ethanol. The mole fraction of water in mixture is (assume ideal behaviour of the mixture) :

• A. 0.1
• B. 0.4
• C. 0.7
• D. 0.9

Answer 1

Answer: (A)

0.1

Weight of $H_{2}O=18g$
Molecular weight of $H_{2}O=18u$
$\therefore$ Moles of $H_{2}O=\frac{18}{18}=1$
Weight of ethanol $\left(C_2{H}_{5}OH\right)=414g$
Molecular weight of $C_2{H}_{5}OH=46u$
$\therefore$ Moles of $C_2{H}_{5}OH=\frac{414}{46}=9$
$\therefore$ Mole fraction of water $=\frac{molesof{H}_{2}O}{molesof{H}_{2}O+molesof{C}_2{H}_{5}OH}$
$=\frac{1}{1+9}$
$\therefore$ Mole of fraction $H_{2}O=0.1$