Question

How would you calculate the vapor pressure of a solution made by dissolving 88.2 g of urea (molar mass = 60.06 g/mol) in 303 mL of water at $35^\circ C$? Vapour pressure of water at 35 degree is 42.18 mmHg.

Answer 1

Hint: To calculate the vapour pressure of the solution we need to determine the mole fraction. The mole fraction of a solvent is calculated by dividing the number of moles of solvent by the total moles present in the solution.

Complete step by step answer:
The mass of urea is 88.2 g.
The molar mass of urea is 60.06 g/mol
The volume of water is 303 mL.
Temperature is $35^\circ C$
The vapour pressure of water is 42.18 mmHg
The urea does not have the vapour pressure, so the vapour pressure of the solution can be determined from the mole fraction of the water and the vapour pressure of the pure water.
The vapour pressure of the solution is calculated by the equation as shown below.
${P_{sol}} = {X_{water}} \times P_{water}^o$
Where,
${P_{sol}}$ is the vapour pressure of the solution.
${X_{water}}$ is the mole fraction of water
$P_{water}^o$ is the vapour pressure of pure water
To calculate the vapour pressure of the solution, first we need to calculate the mole fraction. For knowing the mole fraction, the mass is calculated first using the formula of density.
The density of water is 0.994 g/mL.
The formula of density is shown below.
$D = \dfrac{m}{V}$
D is the density
m is the mass
V is the volume.
To calculate the mass, substitute the values in the above equation.
$ \Rightarrow 0.994g/mL = \dfrac{m}{{303mL}}$
$ \Rightarrow m = 0.994g/mL \times 303mL$
$ \Rightarrow m = 301.2g$
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molecular weight
To calculate the moles of water, substitute the values in the above equation.
$ \Rightarrow n = \dfrac{{301.2g}}{{18g/mol}}$
$ \Rightarrow n = 16.73mol$
To calculate the moles of urea, substitute the values in the above equation.
$ \Rightarrow n = \dfrac{{88.2g}}{{60.06g/mol}}$
$ \Rightarrow n = 1.469mol$
The mole fraction of water is calculated by the formula as shown below.
${X_{water}} = \dfrac{{{n_{water}}}}{{{n_{total}}}}$
Where,
${X_{water}}$ is the mole fraction of water
${n_{water}}$ is the number of moles of water
${n_{total}}$ is the total moles
To calculate the mole fraction of water, substitute the values in the above equation.
$ \Rightarrow {X_{water}} = \dfrac{{16.72mol}}{{16.72 + 1.469mol}}$
$ \Rightarrow {X_{water}} = \dfrac{{16.72mol}}{{18.19mol}}$
$ \Rightarrow {X_{water}} = 0.9192$
To calculate the vapour pressure of the solution, substitute the values in the formula.
$ \Rightarrow {P_{sol}} = 0.9192 \times 42.18mmHg$
$ \Rightarrow {P_{sol}} = 38.8mmHg$
Therefore, the vapor pressure of a solution made by dissolving 88.2 g of urea in 303 mL of water at $35^\circ C$ is 38.8mmHg.

Note:
As the solution is formed by dissolving urea in water, so by general formula to calculate the vapour pressure of the solution the equation should be ${P_{sol}} = {X_{urea}} \times P_{urea}^o + {X_{water}} \times P_{water}^o$but urea is a non-volatile compound so it will not exert any vapour pressure.