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Q. 150g of acetic acid was contaminated with 10.2g ascorbic acid (C6H8O6) to lower down its freezing point by (x×101)C. The value of x is (Nearest integer) [Given Kf=3.9Kkgmol1; Molar mass of ascorbic acid =176gmol1 ]

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Solution:

150gCH3COOH
10.2g ascorbic acid 0.058 moles 
ΔTf=(x×101)C
ΔTf=kf molality 
=3.9×0.058150×1000
=1.5C
=15×101C