A glucose solution which boils at 101.04oC at 1 atm. What will be relative lowering of vapour pressure of an aqueous solutions of urea which is equimolal to given glucose solution? (Given: Kb for water is 0.52 K kg mol-1 )
Answers
Answer:
As we know that, depression in boiling point,
ΔT
b
=k
b
.m
where,
ΔT
b
= Elevation in boiling point = 0.52
k
b
= molal elevation of boiling point = 0.52
m = molality of solution = ?
∴0.52=0.52×m
⇒m=1
As,
m=
Kg of solvent
moles of solute
=1
⇒ As the molality is 1, it signifies that 1 mole of urea is present in 1 kg of water.
∴ no. of moles of water in the solution =
18
1000
=55.55;(∵ 18 is the molar mass of water)
As we know,
mole fraction of solute or solvent in the solution =
total no. of moles of solution
no. of moles of solute or solvent
∴ mole fraction of urea in water =
total no. of moles
no. of moles of urea in the solution
⇒ mole fraction of urea in water =
1+55.55
1
=
56.55
1
=0.0176≈0.018
Hence, the mole fraction of urea in the solution is 0.018.
Hence, the correct option is D
Explanation:
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Answer:
0.034
Explanation:
ΔTb = Kb m ΔTb = 101.04-100 = 1.04 oC
or m= 1.04 /0.52 = 2 m
2 m solution means 2 moles of solute in 1 kg of solvent.
2 m aq solution of urea means 2 moles of urea in 1kg of water.
No. of moles of water = 1000/18 = 55.5
Relative lowering of VP = x2 (where x2 is mole fraction of solute)
Relative lowering of VP = n2/n1+n2 (n2 is no. of moles of solute , n1 is no. of moles
of solvent)
= 2/ 2+55.5 = 2/57.5 = 0.034