`105 mL` of pure water at `4^(circ)C` saturated with `NH_(3)` gas yielded a solution of density `0.9g mL^(-1)` and containing `30%NH_(3)` by mass. Find out the volume of `NH_(3)` solution resulting and the volume of `NH_(3)` gas at `4^(circ)C` and `775 mm` of `Hg`, which was used to saturate water.

Options

A. 66.67 ml
B. 166.67 ml
C. 133.33 ml
D. 266.67 ml

Answer

B

Explanation

Mass of water =105x1=105 g
let wt. of `NH_3` is x
given `x/(x+105)xx100=30`
10x=3x+315
7x=315
x=45
Total mass =45+105=150 g
Volume of solution =mass of solution /density of solution =`150/0.9=1500/9=166.67` ml