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Calculate the amount of KCl which must b...

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

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The correct Answer is:
`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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