Question

$\text{ 6 g }$ of the urea of dissolved in $\text{ 90 g }$ of boiling water. The vapor pressure of the solution is:
A) $\text{ 744}\text{.8 mm }$
B) $\text{ 758 mm }$
C) $\text{ 761 mm }$
D) $\text{ 760 mm }$

Answer 1

the Raoult's law has established a relation between the partial vapour pressure of the volatile component of the solution to the product of vapour pressure of the pure component and the mole fraction of the component. The relation is given as,
$\text{ P = }{{\text{p}}^{\text{0}}}\text{ X }$

Complete step by step answer:
The French scientist F.Raoult studied the vapour pressure of a number of binary solutions of volatile liquid at a constant temperature is known as the Raoult’s law:
The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of the component in the solution.
For a binary solution made of the ‘n’ moles of the liquid and the vapour pressure of the liquid is $\text{ }{{\text{p}}^{\text{0}}}\text{ }$, then according to the Raoult’s law the vapour pressure ‘P’ of the solution can be given as :
$\text{ P = }{{\text{p}}^{\text{0}}}\text{ X }$
Where X is the mole fraction of the component in the solution.
We know that the mole fraction of the component can be given as the ratio of the number of mole of the component to the sum of the number of moles in the solution. The mole fraction is written as,
$\text{ X = }\dfrac{{{\text{n}}_{\text{1}}}}{{{\text{n}}_{\text{1}}}\text{+ }{{\text{n}}_{\text{2}}}}\text{ }$
Where, ${{\text{n}}_{\text{1}}}$ are the number of moles of component 1 and ${{\text{n}}_{\text{2}}}$ are the number of moles of component 2.
We have given the following data.
The weight of urea $\text{ }{{\text{w}}_{\text{1}}}\text{ }$is$\text{ 6 g}$.
The weight of water $\text{ }{{\text{w}}_{2}}\text{ }$is $\text{ 90 g}$
We have given that the solution is prepared in boiling water. The boiling point is the temperature at which the vapour pressure equals the atmospheric pressure. The atmospheric pressure is$\text{ 760 mm of Hg }$.
We are interested in finding the pressure of the solution.
Let’s first calculate the number of moles of water and urea.
$\text{ no}\text{.of moles = }\dfrac{\text{w}}{\text{M}}$
a) Moles of urea:
$\text{ moles of urea (}{{\text{N}}_{\text{Urea}}}\text{)= }\dfrac{\text{w}}{\text{M}}\text{ = }\dfrac{6}{60}\text{ = 0}\text{.1 mole}$
Since the molecular weight of urea is equal to the 60.
b) Moles of water:
$\text{ moles of water (}{{\text{N}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{)= }\dfrac{\text{w}}{\text{M}}\text{ = }\dfrac{90}{18}\text{ = 5}\text{.0 mole}$
Since the molecular weight of water is equal to 18.
The vapour pressure of the solution is calculated as,
$\text{ }{{\text{P}}_{\text{Urea}}}\text{ = }{{\text{p}}^{\text{0}}}\text{ }\dfrac{{{\text{n}}_{\text{Urea}}}}{{{\text{n}}_{\text{Urea}}}\text{+ }{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}}\text{ }$
Substitute the values we have,
$\text{ }{{\text{P}}_{\text{Urea}}}\text{ = (760) }\dfrac{0.1}{\text{0}\text{.1 + 5}\text{.0}}\text{ = 760 }\times 0.0196\text{ = 14}\text{.90 torr}$
We know that, the $\text{ 1 torr = 1 mm of Hg }$
Here we know the pressure of urea . We are interested to find out the vapour pressure of the solution. Therefore, the vapour pressure of the solution is equal to,
$\begin{aligned} & \text{Vapor pressure of solution = Atmospheric pressure }-\text{ Vapor pressure of solute} \\\ & \therefore {{\text{P}}_{\text{solution}}}\text{ = 760 }-\text{ 14}\text{.89 = 745 mm of Hg} \\\ \end{aligned}$
Therefore, here we know the vapour pressure of the solution is equal to the $\text{744}\text{.8 mm of Hg}$.

Hence, (A) is the correct option.

Note: Here, we have taken the partial pressure of the solution as 760 mm of mercury because we know that the boiling point is attained when the vapour pressure of the solution is equal to the atmospheric pressure which is 1 atmosphere or in the height of mercury it is 760 mm of mercury. Note that, the Raoult's law is applicable for the binary solution only. It means the solution needs to have two-component present in the solution. A solution is said to be ideal if it obeys Raoult’s law exactly at all concentrations and at all temperatures.