Question

Find the vapour pressure at the temperature of ${400^ \circ }C$ over $10\%$ aqueous solution of $\left[ {CO{{\left( {N{H_2}} \right)}_2}} \right]$. The vapour pressure of water at ${400^ \circ }C$ is $55.3mmHg$.

Answer 1

Molecular weight of urea is $60g/mol$ and molecular weight of water is $18g/mol$. Now, find the number of moles of urea and number of moles of water then, by finding the mole fraction of water calculate the vapour pressure over aqueous solution.

Complete step by step answer:
The pressure exerted by the vapour in the thermodynamic equilibrium with its condensed phase in a closed system at a given temperature is called the vapour pressure.
Let the temperature be $t$ and vapour pressure be ${p^0}$.
According to the question, it is given that –
$t = {400^ \circ }C \\\ {p^0} = 55.3mmHg \\\$
Let the mass of solution be $100g$.
Therefore, the mass of urea will be $10g$.
Now, calculating the molecular weight of urea $\left[ {CO{{\left( {N{H_2}} \right)}_2}} \right]$ and water ${H_2}O$ -
Molecular weight of urea –
$\left[ {CO{{\left( {N{H_2}} \right)}_2}} \right] = 12 + 16 + \left( {14 + 1 \times 2} \right) \times 2 \\\ \Rightarrow 60g/mol \\\$
Molecular weight of water –
${H_2}O = 2 \times 1 + 16 = 18g/mol$
Now, we have to calculate the number of moles of urea and number of moles of water –
Number of moles of urea, ${X_{urea}} = \dfrac{{10}}{{60}} = \dfrac{1}{6}mol = 0.17mol$
Number of moles of water, ${X_{water}} = \dfrac{{90}}{{18}} = 5mol$
The mole fraction of the compound is given by –

$${x_a} = \dfrac{{{X_a}}}{{{X_a} + {X_b}}} \\\ {x_b} = \dfrac{{{X_b}}}{{{X_a} + {X_b}}} \\\$$

So, the mole fraction of water can be calculated as –
${x_{urea}} = \dfrac{5}{{5 + 0.17}} = \dfrac{5}{{5.17}} \\\ {x_{urea}} = 0.967mol \\\$
Then, the vapour pressure is calculated by using the Raoult’s Law formula –
$p = {p^0} \times {x_{urea}}$
Putting the values of ${p^0}$ and ${x_{urea}}$ in the above formula –
$p = 55.3 \times 0.967 = 53.52mmHg$
Hence, the vapour pressure of urea at temperature of ${400^ \circ }C$ over $10\%$ aqueous solution is $53.52mmHg$.

Note: Raoult’s Law states that partial vapour pressure of each component of the ideal mixture of liquids is equal to the product of vapour pressure of pure component with its mole fraction in the mixture.
When the solution is the volatile component then, according to the Raoult’s law –
$p = p_A^0{x_A}$