Question

A glucose solution which boils at 101.04oC at 1 atm. What will be relative lowering of

vapour pressure of an aqueous solution of urea which is equimolal to given glucose

solution? (Given: Kb for water is 0.52 K kg mol-1
)​

Answer 1

Thanks , I think answer will help you.

Answer 2

The relative lowering of the vapour pressure of an aqueous solution of urea which is equimolal to a given glucose solution is 0.27 atm.

Given:

A glucose solution boils at 101.04oC at 1 atm.

To Find:

The relative lowering of the vapour pressure of an aqueous solution of urea which is equimolal to a given glucose solution.

Solution:

To find the relative lowering of the vapour pressure of an aqueous solution of urea which is equimolal to a given glucose solution we will follow the following steps:

As we know,

∆T = Kbm

Kb is eubilloscopic constant ∆T = T2 - T1 = 101.04-100 = 1.04°C

T2 is the temperature of the solution and T1 is the temperature initially.

m is molality.

So,

$molality \: = \frac{1.04}{0.52} = 2m$

2 molal means 2 moles of glucose in 1 kg of solvent.

Several moles of glucose = Number of moles of urea.

Relative vapour pressure = pi × x1.

Here, pi is the initial pressure, and x1 is the mole fraction of solute.

n1 is several moles of solute while n2 is the number of moles of solvent.

n2 =

$\frac{100}{18} = 5.55 \: moles \: ofsolvent$

x1 =

$\frac{n1}{n1 + n2} = \frac{2}{2 + 5.55} = 0.26$

The vapour pressure of an aqueous solution of urea = pi × x1.

By putting values in the above formula we get,

1 × 0.26 = 0.27

Henceforth, the relative lowering of the vapour pressure of an aqueous solution of urea which is equimolal to a given glucose solution is 0.27 atm.