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CBSE Class 12-science Answered

How many grams of KCL should be added to 1 kg of water to lower its boiling point to -8 degree celsius(kf of water=1.86)?
Asked by lekhakarthikeyan | 25 Aug, 2018, 09:04: PM
answered-by-expert Expert Answer
Given:
 
Weight of Solvent = 1 Kg
 
increment straight T subscript straight f space equals space 8 space to the power of ring operator straight C straight K subscript straight f space space equals space 1.86 We space have comma increment straight T subscript straight f space equals i space straight K subscript straight f space space cross times straight m i space equals space Van apostrophe straight t space Hoff space factor For space KCl space i space equals space 2 increment straight T subscript straight f space equals space 2 cross times 1.86 cross times straight m 8 space equals 3.72 space cross times space straight m straight m space equals space fraction numerator 8 over denominator 3.72 end fraction straight m space equals 2.15 space straight m Molality equals fraction numerator No. space of space moles space of space KCl over denominator Weight space of space Solvent space in space kg end fraction space space space space 2.15 equals fraction numerator No. space of space moles over denominator 1 end fraction No. space of space moles space equals space 2.15 space mol No. space of space moles space equals fraction numerator Weight space of space KCl over denominator Molecular space weight end fraction space space space space space space space 2.15 space space space space space space space equals fraction numerator Weight over denominator 74.55 end fraction Weight space of space KCl space equals space 160.28 space straight g
 
weight of KCl required is 160.28 g
 
 
 
Answered by Varsha | 27 Aug, 2018, 11:36: AM

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