Question

The vapour pressure of pure benzene at 50\[{}^\circ C\] is 268 mm of Hg. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure 167 mm of Hg at 50\[{}^\circ C\]?

Answer 1

Hint: The answer can be found by applying Raoult’s law. Initially, we need to find the mole fraction of solvent and solute by substituting the given values on the mathematical formula of Raoult’s law. From mole fraction of solute, we are going to find out the number of moles of solute.

Complete step by step solution:
 This problem is a direct application of Raoult’s law. Raoult’s law for non-volatile solute states that the relative lowering of the vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of solute in the solution. This relation can be mathematically represented as follows,
\[{{X}_{2}}\]= $\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$ = \[{{X}_{2}}\]
where \[{{X}_{2}}\] is the mole fraction of solute
\[{{n}_{1}}\] are the moles of solvent
\[{{n}_{2}}\] are the moles of solute
\[{{P}^{0}}\] is the vapour pressure of solution (Given as 167 mm of Hg)
\[{{P}_{s}}\] is the vapour pressure of pure solvent (Given as 268 mm of Hg)
Raoult's law can also be written more simply and this form of the equation is the one we are going to use to solve the problem. The relation is as follows
\[{{P}_{s}}\] = Mole fraction of the solvent (\[{{X}_{2}}\]) × \[{{P}^{0}}\]
The mole fraction of the solvent can be found by rearranging the above equation,
Mole fraction of solvent = \[\dfrac{{{P}_{s}}}{{{P}^{0}}}\]
By substituting the values of \[{{P}_{s}}\] and \[{{P}^{0}}\] on the above equation we get,
Mole fraction of solvent = \[\dfrac{167}{268}\]
=0.623
By subtracting the value of mole fraction of solvent from 1 we can find the mole fraction of solute since the sum of mole fractions of solute and solvent will always be 1.
Thus, the mole fraction of non-volatile solute = 1−0.623
                                                                                   = 0.377
We are asked to find the number of moles of benzene. It can be found from the ratio between the mole fraction of solute and solvent. It can be written as
\[\dfrac{n}{N}=\dfrac{Mole\quad fraction\quad of\quad solute}{Mole\quad fraction\quad of\quad solvent}=\dfrac{0.377}{0.623}\]

= 0.605 moles

The number of moles of non-volatile benzene required for lowering of vapour pressure is 0.605 moles.

Note: Since there are Raoult’s laws for both volatile and non-volatile solutes, it's important to pay attention to the question to find whether volatile or non-volatile solute is given. Another chance of possible error is when we stop with finding mole fraction. We are asked to find the number of moles, not mole fraction.