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)

Solution

straight W subscript straight B space equals ? straight W subscript straight A space equals 1000 straight g increment straight T subscript straight f space equals 2 space straight K straight K subscript straight f space equals 1.86 space kg space mol to the power of negative 1 end exponent straight M subscript straight B space of space KCl space equals space 39.09 space plus 35.45 space equals 74.54 space mol to the power of negative 1 end exponent increment straight T subscript straight f equals space fraction numerator straight K subscript straight f space straight x space straight W subscript straight b space straight x space 1000 over denominator straight W subscript straight A space straight x space straight M subscript straight B end fraction equals space fraction numerator 1.86 space straight x space straight W subscript straight B space straight x space 1000 over denominator 1000 space straight x space 74.54 end fraction straight W subscript straight B space equals space fraction numerator 2 space straight x space 74.54 over denominator 1.86 end fraction straight W subscript straight B space equals space 80.15 space straight g

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