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CBSE Class 12-science Answered

Density of 1M of a non-electrolyte aqueous solute of molecular weight 100 is 2 g/mL. If Kf(H2O) is 1.86°C kg mol–1, then solution will freeze at
Asked by Yogesh | 05 Aug, 2018, 12:11: PM
answered-by-expert Expert Answer
Given:
 
Density = 2 g/mL.
 
Kf = 1.86°C kg/mol
Molarity = 1M
 
Molecular weight =100
 
freezing point =?
 
 
 
Volume space of space solution space equals space 1000 ml Mass space of space solution space equals space Volume space cross times space Density space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1000 cross times space 2 Mass space of space solution space equals space 2000 space straight g Amount space of space solute space in space 1000 space ml space of space solution space equals space 1 cross times space Molecular space weight space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 100 space straight g Mass space of space water space equals Mass space of space solution minus space Mass space of space solute space space space space space space space space space space space space space space space space space space space space space space equals space 2000 minus 100 Mass space of space water space equals 1900 space gm Molality space equals fraction numerator No. space of space moles space of space solute over denominator Weight space of space solvent space in space Kg end fraction space space space space space space space space space space space space equals fraction numerator 1 over denominator begin display style bevelled 1900 over 1000 end style end fraction Molality equals 0.5263 space straight m We space know comma increment straight T subscript straight f space equals space straight K subscript straight f times straight m space space space space space space space equals space 1.86 cross times 0.5263 increment straight T subscript straight f space space equals space 0.9789 space to the power of ring operator straight C increment straight T subscript straight f space equals straight T to the power of 0 subscript straight f minus straight T subscript straight f straight T subscript straight f space equals space straight T to the power of 0 subscript straight f minus increment straight T subscript straight f space space space space space equals 0 minus space 0.9789 straight T subscript straight f space space equals negative space 0.9789 space to the power of ring operator straight C space space space space space equals negative space 0.9789 plus 273 space space space straight T subscript straight f equals space 272.02 space straight K
 
Freezing point is 272 K
Answered by Varsha | 06 Aug, 2018, 01:29: PM

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