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Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_(f) " for water = " 1.86 " K kg "mol^(-1)) |
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Answer» Solution :SINCE one mole of KCI gives 2 mole PARTICLES, the value of I = 2 `DELTA T_(f) = 2 K ` `K_(f) = 1.86 " kg mol"^(-1)` Applying equation, `T_(f) = iK_(f) `m `m = (Delta T_(f))/(iK_(f)) = (2)/(2 xx 1.86)` Therefore, 0.54 mole of KCI should be added to one kg of water. MOLAR mass of KCI = 39 + 35.5 = 74.5 g AMOUNT of KCl = 0.54 `xx 74.5 `g = 40.23 g |
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