Solutions Question 70

Question: The molar freezing point constant for water is $ {{1.86}^{o}}C,mol{e^{-1}} $ . If 342 gm of canesugar $ (C_{12}H_{22}O_{11}) $ are dissolved in 1000 $ gm $ of water, the solution will freeze at [NCERT 1977; CPMT 1989; Roorkee 2000; DCE 2004]

Options:

A) $ -{{1.86}^{o}}C $

B) $ {{1.86}^{o}}C $

C) $ -{{3.92}^{o}}C $

D) $ {{2.42}^{o}}C $

Show Answer

Answer:

Correct Answer: A

Solution:

$ \Delta T_{f}=1.86\times ( \frac{342}{342} )={{1.86}^{o}} $ ; \ $ T_{f}=-{{1.86}^{o}}C $ .



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