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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)`). |
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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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