Question

Laboratory ammonia is $15.2M$ $N{H}_3$ (aqueous) with a density of $0.8980g/mL$. Mole fraction of $N{H}_3$ in this solution is

• A. $0.299$
• B. $0.708$
• C. $0.148$
• D. $0.858$

Answer 1

The correct option is A $0.299$

$15.2M$ $N{H}_3$ (aqueous) means $15.2mol$ of $N{H}_3$ in $1000mL$ solution.
$\text{1000 mL solution}=1000\times 0.898g=898g\text{solution}$
$=898g\text{solution}(N{H}_3+H_{2}O)$
$15.2molN{H}_3=15.2\times 17gN{H}_3$
$=258.4gN{H}_3$
Thus, $H_{2}O$ in solution $=898.0-258.4=639.6g$
$\text{moles of}{H}_{2}O=\frac{639.6}{18}mol=35.5mol$
$n\left(H_{2}O\right)+n\left(N{H}_3\right)=35.5+15.2=50.71=50.7mol$
$\therefore {\chi }_{N{H}_3}=\frac{15.2}{50.7}=0.299$