A sea water sample has density of `1.03g//cm^(3)` and `2.8% NaCl` by mass. A saturated solution of `NaCl` in water is `5.45 M NaCl`. How much water would have to be evaporated from `10^(6)` litres of sea water before `NaCl` would precipitate?

Answer

`9.095xx10^(5)L`

Explanation

`D=1.03g//cm^(3)`
`2.8%NaClrarr100g` solution contain `2.8g NaCl`
`V=(100)/(1.03xx1000)L`
`1Lrarr(2.8xx10.3)/(58.5)=(0.493)`
`M_(2)v_(2) = " " M_(1)V_(1)`
` 0.493xx10^(6)=5.45xxV_(1)`
`V_(1)=9 xx 10^(4)`
so water evaporated `=10^(6)-9xx10^(4)=9.095xx10^(5)L`