Question

The vapour pressure of pure water is 23.5 mm Hg. Then, the vapour pressure of an aqueous solution which contains 5 mass percent of urea is:
(Molar mass of urea is 60)
A.23 mm Hg
B.18 mm Hg
C.31 mm Hg
D.35 mm Hg

Answer 1

Hint : We must remember that, with the addition of any non-volatile solute in solvent the vapour pressure of the solvent decreases.
Formula used: pressure (P) = Mole fraction of solvent × vapour pressure of pure solvent

Complete step by step solution :
Molar mass of urea is 60
The vapour pressure of pure water is $23.5{\text{ }}mm{\text{ }}Hg$
Complete step to step solution:
We are given the following information:
Vapour pressure of pure water is $23.5{\text{ }}mm{\text{ }}Hg$
If total mass of the solution 100g then, the mass of urea is
Mass of urea = 5g (as aqueous solution contains 5 mass percent of urea)
Similarly, the mass of water will be 100g – 5g = 95g.
Now, we will calculate the moles of urea and water
Number of moles of urea present in aqueous solution = $\dfrac{{{\text{Given Mass}}}}{{{\text{Molecular Mass}}}}$=$\dfrac{{\text{5}}}{{60}} = 0.083$.
Similarly, Number of moles of water present in aqueous solution = $\dfrac{{{\text{Given Mass}}}}{{{\text{Molecular Mass}}}}$=$\dfrac{{95}}{{18}} = 5.278$.
So, the total number of moles present in solution = Moles of urea + moles of water
$= {\text{ }}0.083 + 5.278{\text{ }} = {\text{ }}5.361.$
Now, we have to calculate the mole fraction of solvent (water) which will be
Mole fraction of solvent (Water) = $\dfrac{{5.278}}{{5.361}}$
We know that the pressure (P) = Mole fraction of solvent ×vapour pressure of pure solvent
Pressure (P) = $\dfrac{{5.278}}{{5.361}} \times 23.5mm{\text{ }}Hg$
So, the pressure (P) is equal to $23.14{\text{ }}mm{\text{ }}Hg$ which is approximately equal to$23mm{\text{ }}Hg$.
So the vapour pressure exerted by aqueous solution of 5 mass percent urea is $23mm{\text{ }}Hg$.
Therefore, we can conclude that the answer for the question is option A.

Note : We must remember that the Vapour pressure is a tendency of a material to change into a gaseous or vapour state and it increases as temperature increases. And also vapour pressure on a liquid’s surface becomes equal to the surrounding pressure. We call it a liquid's boiling point.