Question

105 ml of pure water at $4^{\circ }$C is saturated with $N{H}_3$(g) producing a solution of density 0.9 gm/ml. If this Contain 30% of $N{H}_3$ by mass, the total volume of solution is -

• A. 250 ml
• B. 125 ml
• C. 166.67 ml
• D. 111.11 ml

Answer 1

The correct option is C 166.67 ml

Let. final vol. of solution is V ml
$\therefore$ wt of soln = V $\times 0.9$gm and wt of $N{H}_3$
= $V\times 0.9-105$ gm
$\therefore V\times 0.9-105=V\times 0.9\times 0.3$
$\therefore V\times 0.9\times 0.7=105$

Or
$V=\frac{105}{0.9\times 0.7}=\frac{105}{0.63}$ ml.
= 166.67 ml