What weight of sucrose will be dissolved in 250 g of water and obtain solution with boiling point at 100.2 degree Celsius
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Answer:
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Explanation:
Let the molality of the solution is m.
ΔT
b
=
K
f
b×
m
ΔT
f
= K
f
× m
ΔT
f
+ΔT
b
=
(K
f
+
K
b
)×
m=
(1.86+
0.51)×
m
=
2.37×
m
Hence
m=
5/2.37=
2.11
and
1000
2.11×100
=
0.211
gram of sucrose must be dissolved in 100 gram of water.
Required mass of sucrose
=
0.211mol×
342g/mole=
72.2g
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