1.

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)

Answer»

`0.24 dm^(3)`
`0.12 dm^(3)`
`1.2 dm^(3)`
`0.2 dm^(3)`

Solution :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)`


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