`NaCl` is added to`1` litre water to such an extent that `DeltaT_(f)//K_(f)` becomes to `(1)/(500)`,wt.of `NaCl` added to

Options

A. `5.85g`
B. `0.585g`
C. `0.0585g`
D. `0.0855g`

Answer

C

Explanation

For `NaCl i=2`
or`DeltaT_(f)=i.K_(f).m`
`DeltaT_(f)//K_(f)=2.m.`
Or =`2m`
Or molarity `=(1)/(1000)or(1)/(1000)` moles are dissolved `Kg^(-1)`of water or approximately `(1)/(1000)`moles are present in `1.0` litre of water(considering molarity and molality to be the same)
`:.` Weight of `NaCl=(1)/(1000)xx58.5 g` of `NaCl`
`= 0.0585 g` of NaCl.