Question

How much urea (molar mass \[60g/mol\]) should be dissolved in \[50g\] of water so that its vapour pressure at room temperature is reduced by \[25\% \]. [Write up to 1 decimal place]

Answer 1

Hint: As we know that the urea is a chemical compound having the chemical formula, \[CO{\left( {N{H_2}} \right)_2}\]. The urea contains carbon, nitrogen and oxygen atoms. And the urea is also known as carbamide which contains two amino groups. This amino group is joined with the carbonyl functional group. The molar mass of urea is equal to \[60g/mol\]. And water is a chemical compound having the symbol \[{H_2}O\] with molecular mass \[18g\].

Complete answer:
The weight of the urea can be find out by using the equation,
$\dfrac{{P_1^0 - {P_1}}}{{P_1^0}} = \dfrac{{{W_1} \times {M_2}}}{{{W_2} \times {M_1}}}$ _____ (1)
Where, \[P_1^o\] is the vapour pressure of the pure solvent, X.
\[{P_1}\]is equal to the vapour pressure of the solution of urea. Which implies,
\[P_1^o - {P_1} = 25\% \] of \[x = 18.02g/mol\] and \[{W_1}\]is equal to the weight of water which is equal to \[50g\]. \[{M_2}\]is equal to the molar mass of urea which is equal to \[60g/mol\].
Substitute the values in the equation one, will get the weight of phenol dissolved in \[50g\] of water;
$\dfrac{{0.25x}}{x} = \dfrac{{{W_2} \times 18.02}}{{50 \times 60}}$
By rearranging the equation,
${W_2} = \dfrac{{0.25 \times 3000}}{{18.02}} = 41.62g \simeq 41.7g$
Therefore, the weight of phenol dissolved in \[50g\] of water is equal to \[41.7g\].

Note:
We have to remember that the elevation in boiling point explains the method that the boiling point of a liquid will be more by adding another compound. And the vapour pressure is also known as equilibrium vapour pressure which is equal to the exerted pressure by a vapour in the equilibrium in thermodynamics in a closed system with its condensed phases at a particular temperature. And the weight of the solution can be found by using the equation of vapour pressure.