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 $ .
