Question

A solution contain $36\%$ water and $64\%$ acetaldehyde $\left(C{H}_{2}CHO\right)$ by mass. The mole fraction of acetaldehyde is:

• A. $0.42$
• B. $0.2$
• C. $4.2$
• D. $2.1$

Answer 1

The correct option is A $0.42$

$36$% $H_{2}O$

$36g$ of $H_{2}O$ in $100g$ of solution

$64$% ${CH}_{3}CHO$(acetaldehyde)

$64g$ of ${CH}_{3}CHO$ in $100g$ of solution

Mole fraction$\left(X_A\right)=\frac{n_A}{n_A+n_B}{n}_A=\frac{64g}{44g}{n}_B=\frac{32g}{18g}$

Acetaldehyde

$X_A=\frac{64/44}{64/44+32/18}=\frac{1.45}{1.45+1.77}=\frac{1.45}{3.22}=0.45$

Mole fraction of acetaldehyde is $0.45$