Question

A solution containing \[30\,gm\] of non-volatile solute exactly in \[90\,gm\] of water has a vapour pressure of \[2.8\,kPa\] at \[298\,K\] . Further, \[18\,gm\] of water is then added to the solution and the new vapour becomes \[2.9\,kPa\]at \[298\,K\].
Calculate:
(i) Molar mass of the solute
(ii) Vapour pressure of water at \[298\,K\]

Answer 1

Hint: Chemical name for $KMn{O_4}$ is Potassium Permanganate. It is a very good oxidising agent. It persists a purple colour and is used in titrations. Due to its colour, it acts as a self-indicator in titrations. Oxalic acid is an organic acid in which two carboxylic acid group are joined together and it has a chemical formula as ${(COOH)_2}.$

Complete answer:
According to Raoult's law, the partial vapour pressure of a solvent in a solution is equal to the vapour pressure of the pure solvent multiplied by the mole fraction in the solution or we can say it is equivalent to the vapor pressure of the pure solvent measured in the solution by the mole fraction.
Formula for Raoult’s law:
\[\dfrac{{{p^o} - p}}{{{p^o}}} = \dfrac{{w\, \times \,M}}{{m\, \times \,W}}\]
Where,
\[{p^o}\] is the vapour pressure of the water
\[p\] is the vapour pressure of the solution
\[M\] is the molecular weight of the solvent
\[W\] is the weight of the solvent
\[w\] is the weight of the solute
\[m\] is the molecular weight of solute
In this question, it is given,
Weight of solute, \[w\]= \[30\,gm\]
Weight of water, \[W\]= \[90\,gm\]
Vapour pressure of the solution, \[p\]= \[2.8\,kPa\]
Molecular weight of the water, \[M\]= \[18g/mol\]
Substituting all these values in Raoult’s law formula, we get
\[\dfrac{{{p^o} - 2.8}}{{{p^o}}} = \dfrac{{30\, \times \,18}}{{m\, \times \,90}}\]
\[\dfrac{{{p^o} - 2.8}}{{{p^o}}} = \dfrac{{540}}{{m\, \times \,90}}\]
Now, solve the Right hand side, we get
\[\dfrac{{{p^o} - 2.8}}{{{p^o}}} = \dfrac{6}{m}\]
By cross multiplying, we get
\[m({p^o} - 2.8) = 6{p^o}\]
\[m{p^o} - 2.8m = 6{p^o}\]
Rearranging both sides, we get
\[m{p^o} - 6{p^o} = 2.8\]
Take common \[{p^o}\] outside on left hand side,
\[{p^o}(m - 6) = 2.8\]
\[\dfrac{{2.8}}{{{p^o}}} = m - 6\]…… (1)
After adding water,
Weight of solute, \[w\]= \[30\,gm\]
Weight of water, \[W\]= \[90\,gm\, + 18gm\]=\[108gm\]
Vapour pressure of the solution, \[p\]= \[2.9\,kPa\]
Again substitute all the values in the above formula, we get,
\[\dfrac{{{p^o} - 2.9}}{{{p^o}}} = \dfrac{{30\, \times \,18}}{{m\, \times \,108}}\]
\[\dfrac{{{p^o} - 2.9}}{{{p^o}}} = \dfrac{{540}}{{m\, \times \,108}}\]
Now, solve the Right hand side, we get
\[\dfrac{{{p^o} - 2.9}}{{{p^o}}} = \dfrac{5}{{m\,}}\]
By cross multiplying, we get
\[m({p^o} - 2.9) = 5{p^o}\]
\[m{p^o} - 2.9m = 5{p^o}\]
Rearranging both sides, we get
\[m{p^o} - 5{p^o} = 2.9\]
Take common \[{p^o}\]outside on left hand side
\[{p^o}(m - 5) = 2.9\]
\[\dfrac{{2.9}}{{{p^o}}} = m - 5\]…… (2)
Divide equation (1) by (2), we get
\[\dfrac{{2.8}}{{2.9}} = \dfrac{{m - 6}}{{m - 5\,}}\]
By cross multiplying, we get
\[2.8\,(m - 5) = 2.9\,(m - 6)\]
\[2.8m - 14 = 2.9m - 17.4\]
Rearranging both sides, we get
\[2.8m - 2.9m = - 17.4 + 14\]
\[ - 0.1m = - 3.4\]
\[m = \dfrac{{ - 3.4}}{{ - 0.1}}\]
 m = \[34g/mol\]
The molar mass of the solute (m) is \[34g/mol\]
Now, substitute the values of m = \[34g/mol\] in equation (1) we get,
\[\dfrac{{2.8}}{{{p^o}}} = \dfrac{{34 - 6}}{{34\,}}\]
\[\dfrac{{2.8}}{{{p^o}}} = \dfrac{{28}}{{34\,}}\]
By solving, we get
\[{p^o} = \dfrac{{2.8 \times 34}}{{28}}\]
\[{p^o} = \dfrac{{95.2}}{{28}}\]
So, we get the value of \[{p^o}\]
\[{p^o}\, = \,3.4\,kPa\]
So, the vapour pressure of water at \[298\,K\]is \[{p^o}\, = \,3.4\,kPa\] and the molar mass of the solute (m) is \[34g/mol\].

Note:
Molecular weight of the substance is also known as molar mass. It is the measure of the total mass of the atoms in grams which forms a molecule. The unit for the measurement for the molar mass of the substances is gram per mole.