1.

Calculate the amount of `KCl` which must be added to `1kg` of water so that the freezing point is depressed by `2K`. (`K_(f)` for water `=1.86K kg "mol"^(-1)`).

Answer» Correct Answer - `KCIrarrK^(+)+Cl^(-)rArri=2`
`DeltaT_(f)=iK_(f)xxm`
`2K=2xx1.86xxm rArrm=(1)/(1.86)`
`m=0.5376 "mol"//kg`
Amount of `KCl=0.5376xx74.5=40.05g`.


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