JEE Main 2022 Chemistry question
150 g of acetic acid was contaminated with 10.2 g ascorbic acid to lower down its freezing point by . The value of x is . (Nearest integer) (Given : = 3.9 K kg ; molar mass of ascorbic acid = 176 g )
Answer: 15
Solution
= (10.2/176)/(0.150) =
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