105 ml of pure water at 4° C is saturated with NH 3( g ) producing a solution of density 0.9 g / ml. If this solution contains 30 % of NH 3 by mass, therefore the total volume of solution is:

Question

105 ml of pure water at 4° C is saturated with NH 3( g ) producing a solution of density 0.9 g / ml. If this solution contains 30 % of NH 3 by mass, therefore the total volume of solution is: (A) 250 ml (B) 125 ml (C) 166.67 ml (D) 111.11 ml

Tardigrade Admin · Accepted Answer

Let final vol. of solution be V ml ∴ Weight of solution =V × 0.9 g and wt . of NH 3 =(V × 0.9-105) g ∴ V × 0.9-105=V × 0.9 × 0.3 ∴ V × 0.9 × 0.7=105 Or V=(105/0.9 × 0.7)=(105/0.63) ml =166.67 ml

Q. $105 m l$ of pure water at $4^{\circ} C$ is saturated with $N H_{3} (g)$ producing a solution of density $0.9 g / m l$ . If this solution contains $30%$ of $N H_{3}$ by mass, therefore the total volume of solution is:

250 ml

125 ml

166.67 ml

111.11 ml

Let final vol. of solution be $V m l$

$∴$ Weight of solution $= V \times 0.9 g$ and $wt .$ of $N H_{3}$

$= (V \times 0.9 - 105) g$

$∴ V \times 0.9 - 105 = V \times 0.9 \times 0.3$

$∴ V \times 0.9 \times 0.7 = 105$

Or $V = \frac{105}{0.9 \times 0.7} = \frac{105}{0.63} m l = 166.67 m l$

Q. 105ml of pure water at 4∘C is saturated with NH3​(g) producing a solution of density 0.9g/ml. If this solution contains 30% of NH3​ by mass, therefore the total volume of solution is: