Question

Mole fraction of ethyl alcohol in aqueous ethyl alcohol $\left(C_2{H}_{5}OH\right)$ solution is $0.25$. Hence percentage of ethyl alcohol by weight is:

• A. $54\%$
• B. $25\%$
• C. $75\%$
• D. $46\%$

Answer 1

Answer: (D)

$46\%$

If ethanol is 0.25, then H${}_2$O will be 0.75.

Mol wt of ethanol is 46.06 g/mol,

So, 0.25 mol=11.515g of ethanol

Mol wt of H${}_2$O is 18.01528g/mol,

So, 0.75 mol=13.51g of H${}_2$O

Total wt=25.02646 g

wt % of etanol is $=\frac{11.515}{25.02646}\times 100=46.011\approx 46$%

Hence, the correct option is $D$