Question

At $30^\circ$C, pure benzene (molecular weight 78.1) has a vapour pressure of 121.8 mm. Dissolving 15.0 g of a non-volatile solute in 250 g of benzene produced a solution having vapour pressure of 120.2 mm. Determine the approximate molecular weight of the solute.
a.) 152.3
b.) 355.8
c.) 392.4
d.) 409.2

Answer 1

Hint: According to the relative lowering of vapour pressure when we dissolve non-volatile solute in a pure solvent the vapour pressure of the pure solvent gradually decreases.
Relative lowering of vapour pressure = \[\dfrac{P-\,{{P}_{s}}}{P}\]

Complete step by step solution:
Suppose P be the vapour pressure of pure solvent and \[{{P}_{s}}\] be the vapour pressure of the solution i.e. when the non- volatile solute is added to the solvent, we can write the lowering in vapour pressure as (\[P-\,{{P}_{s}}\]). Now this vapour pressure is relative to the vapour pressure of pure solvent.
Therefore, the relative lowering in vapour pressure = \[\dfrac{P-\,{{P}_{s}}}{P}\]
Now according to Raoult's law the vapour pressure of a pure solvent is caused by the number of molecules evaporating from the surface of the solvent. So, when we add a non-volatile solute to the solvent the escape of the solvent particles from the surface decreases, Thus, lowering of vapour pressure takes place.
The relation between Raoult’s law and the relative lowering of vapour pressure =
\[\dfrac{P-\,{{P}_{s}}}{P}\] = \[\dfrac{{{n}_{2}}}{{{n}_{1}}\,+\,{{n}_{2}}}\] (i)
Where, \[{{n}_{1}}\] = number of moles of solvent.
\[{{n}_{2}}\] = number of moles of solute.
Now, according to the question, we have to find the molecular weight of the solute. So, let’s take it x.
Pure benzene is the solvent and its vapour pressure (P) = 121.8mm
Vapour pressure of the solution (\[{{P}_{s}}\]) = 120.2
\[{{n}_{1}}\] = given weight of solvent / molecular weight of solvent
\[{{n}_{1}}\] = 250/ 78.1
\[{{n}_{1}}\] = 3.2
\[{{n}_{2}}\] = given weight of solute / molecular weight of solute.
\[{{n}_{2}}\] = 15/x
Now putting the values in equation (i)
\[\dfrac{121.8-\,120.2}{121.8}\,=\,\dfrac{15/x}{15/x\,+\,3.2}\]
\[\dfrac{1.6}{121.8}\,=\,\dfrac{15}{15+\,3.2x}\]
\[0.013=\,\dfrac{15}{15+\,3.2x}\]
0.013 (3.2x + 15) = 15
0.0416x + 0.195 = 15
0.0416x = 14.805
x = 14.805/ 0.0416
x = 355.8

Therefore, the correct answer to the question is (b).

Note: Deviations from Raoult’s law:
Positive deviation, when the interactions between the molecules of the solution is weaker than the interactions between the molecules of the solvent.
Negative deviation, when the interactions between the molecules of the solution is stronger than the interactions between the molecules of the solvent.