Question

A solution containing 34.2g of cane sugar dissolve in 500 cm^3 of water froze at -0.374°C. calculate freezing point

Answer 1

Answer:

Explanation:

A solution of sucrose (molar mass = 342 g mol  

−1

) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be:  

[K  

f

​  

 for water =1.86 K kg mol  

−1

]

December 30, 2019

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Charu Sravani

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ANSWER

DepressioninFreezingpointΔT  

f

​  

=  

M  

1

​  

×W  

2

​  

 

W  

1

​  

×K  

f

​  

×1000

​  

 

whereW  

1

​  

=Weight of Solute

W  

2

​  

=Weight  of solvent

M  

1

​  

=Molar mass of solute

K  

f

​  

=Freezing point deprssion constant

Now,ΔT  

f

​  

=  

342×1000

1.86×68.5×1000

​  

=0.372C

Now,ΔT  

f  =T  o  −T  f

​So,T  f

​  

=0−0.372=−0.372C.

(Freezing point of purewater=0  

0

C.)

Hence, the correct option is A