Solutions Question 70

Question: The molar freezing point constant for water is 1.86oC,mole−1 . If 342 gm of canesugar (C12H22O11) are dissolved in 1000 gm of water, the solution will freeze at [NCERT 1977; CPMT 1989; Roorkee 2000; DCE 2004]

Options:

A) −1.86oC

B) 1.86oC

C) −3.92oC

D) 2.42oC

Show Answer

Answer:

Correct Answer: A

Solution:

ΔTf=1.86×(342342)=1.86o ; \ Tf=−1.86oC .

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