The wt. percent of `NO_(3)^(-)` in the solute in a solution is 20. The volume of solvent containing 60 g of solute (d = 1.2 g/cc)

Options

A. `0.24 dm^(3)`
B. `0.12 dm^(3)`
C. `1.2 dm^(3)`
D. `0.2 dm^(3)`

Answer

D

Explanation

Wt. of solvent = Wt. of solution- Wt. of solute
`=100 - 20 =80g`
20 g of solute is present in 80 g of solvent.
60 g of solute will be present in `(80 xx 60)/20=240 g
`:."Volume of solvent" = (Wt)/"density"=240/1.2=200 cm ^(3)`
= `0.2 dm^(3)`