Question

When 0.1 mole of sugar is dissolved in
200g of water the solution boils under a
pressure of 1 atm at
(kg = 0.513 K. kg mol-')
(A) 100.513
(B) 100.0513
(C) 100.256
(D) 101.025

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Answer 1

Answer:

c) 100.256 °C

Explanation:

First we need to calculate DeltaTb

DeltaTb = (1000*w2*Kb)/(w1*M2)

molecular weight of sugar (C12H22O11) = 12*12+1*22+16*11 = 342g/mol which is M2 (molecular weight of salute)

M2 = 342g/mol

1 mole of sugar is 342g

0.1mole is how much....?

= 0.1*342

= 34.2g which is w2

w1 is given = 200g

boiling temperature of water Tbo = 373.15 °K

Kb is given = 0.513 K-Kg mol-1

now,

DeltaTb = (1000*34.2*0.513)/(200*342)

DeltaTb = 17544.6/68400

DeltaTb = 0.256

DeltaTb = Tb - Tbo

0.256 = Tb - 373.15

Tb = 373.406 °K

in degree centigrade,

Tb = 373.406 - 273.15

= 100.256 °C

Answer is option C i.e. 100.256 °C