Question

an aqueous solution of urea is found to boil at 100.52°C given for water is 0.52kkgmol-1 the mole fraction of urea in solution is?​

Answer 1

Explanation:

it means one mole of urea is dissolved in one of(55.5moles)of water then mole fraction is

no. of moles of urea/total moles of mixture

1/56.5=.0177

Answer 2

Answer:

The urea's mole fraction in the solution measured is $0.017$.

Explanation:

Given,

The molal elevation constant, $K_{b}$ = $0.52K kgmol^{-1}$

The boiling temperature, $T_{b}$ = $100\textdegree C$

The mole fraction of urea =?

Firstly, we have to calculate the molality from the relation given below:

• $\Delta T = K_{b}\times m$    -------equation (1)

Here,

• $\Delta T$ = The change in the boiling point
• $K_{b}$ = The molal elevation constant
• m = Molality

Also,

• $\Delta T$ = $T_{b} - T$

Here,

• T = The boiling temperature of water = $100\textdegree C$

Therefore,

• $\Delta T = 100.52 - 100$ = $0.52\textdegree C$

After putting this value in equation (1), we get:

• $0.52 = 0.52\times m$
• m = $1 molal$

As we know,

• $1 molal$ means $1 mole$ of solute on $1000g$ of solvent.

Therefore, there is one mole of urea ($n_{u}$).

As given, the solvent is water and the mass of the solvent is $1000g$ as calculated above.

Also, the molar mass of water = $18g/mol$

Now,

• The number of moles of water, $n_{w}$ = $\frac{Mass}{Molar mass}$ = $\frac{1000}{18}$ = $55.55mol$

Thus,

• The mole fraction of urea = $\frac{n_{u} }{n_{u}+n_{w} }$ = $\frac{1}{1+ 55.55}$ = $\frac{1}{56.55}$ = $0.017$

Hence, the urea's mole fraction in the solution = $0.017$.