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