Question

An aqueous solution containing 288 gm of a non- volatile compound having the
stochiometric composition CxH2xOx in 90 gm water boils at 101.24°C at 1.00 atmospheric
pressure. What is the molecular formula?
Ko(H2O) = 0.512 K mol-' kg, Tb (H20) = 100°C.​

Answer 1

Elevation in BP = 101.24-100 = 1.24℃

Since solution is non-volatile,

Elevation in BP is :

∆T = k × i × w/m × 1000/W

1.24 = 0.512 × × 288/m × 1000/90

Therefore, m = 1321.2 gm/mol

m = molar weight of solution

Molar Mass(CxH2xOx) = 12x+2x+16x = 30x

Therefore, 1321.2 = 30x

Therefore, x = 44

Hence, molecular formula is C44H88O44.

Answer 2

Answer

The elevation in the boiling point ΔT

b

=101.24−100=1.24

o

C

ΔT

b

=K

b

m

1.24=0.512m

m=2.422

Here m is the molality

The molality of the solution m=

molar mass of solute×mass of solvent (in kg)

mass of solute

=

M×0.090

288

The molar mass of solute M=1321g/mol

From molecular formula C

x

H

2x

O

x

the molar mass is 12x+2x+16x=30x

Hence, 30x=1321

x=44

Hence, x+2x+x=4x=4(44)=176

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