Question

6g. of urea is dissolved in 90g. of boiling water. The vapour pressure of the solution is

Answer 1

This is the initial formula of vapour pressure.....
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Answer 2

To calculate the vapour pressure of the solution when 6g.  

Urea is dissolved in 90g.  

Boiling water the following information are required:

Molar weight of Urea = 60

Number of moles in 6g of Urea $=\frac{6}{60}=0.1\ moles$

Molar weight of water = 18

Number of moles in 90g of water $=\frac{90}{18}=5\ moles$

Molar fraction of the solution = Moles of water / (total number of moles of solution)

$=\frac{5}{(5+0.1)}=\frac{5}{5.1}=0.98\ \mathrm{atm}$

Since the temperature is given to be the boiling point of water = 100 degree Celsius, therefore, pressure $\left(\mathrm{P}_{0}\right)=1\ atmosphere$

Vapour Pressure of the solution $\begin{aligned}=& \mathrm{P}_{0} \times \text { Molar Fraction }=1 \times 0.98 \text { atm. } \\ &=0.98 \text { atm. } \end{aligned}.$