Question

The vapour pressure of benzene at ${\text{90}}{\,^{\text{o}}}{\text{C}}$ is ${\text{1020}}$ torr. A solution of $5$ g of a solute in $58.5$g benzene has vapour pressure $990$torr. The molecular mass of the solute is:
A. $78.2$
B. $178.2$
C. $206.2$
D. $220$

Answer 1

Hint: To answer this question we should know what is lowering of vapour pressure and its formula. Relative lowering of vapour pressure relates the molecular mass of solvent and solute with vapour pressure. First we will write the formula of relative lowering of vapour pressure we will substitute the given data and calculate the molecular mass of solvent.


Complete solution:
The formula of the relative lowering of vapour pressure is as follows:
$\dfrac{{{{\text{p}}_{\text{o}}}\, - {{\text{p}}_{\text{1}}}}}{{{{\text{p}}_{\text{o}}}}}{\text{ = }}\dfrac{{{{\text{w}}_2} \times {{\text{M}}_1}}}{{{{\text{M}}_2}\, \times {{\text{w}}_1}}}$
Where,
${{\text{p}}_{\text{o}}}$is the vapour pressure of pure solvent
${{\text{p}}_{\text{1}}}$ is the vapour pressure of solution
${{\text{w}}_1}$ is the mass of solvent.
${{\text{w}}_2}$ is the mass of solute.
${{\text{M}}_1}$ is the molar mass of the solvent
${{\text{M}}_2}$ is the molar mass of the solute
Here, the solvent is benzene.
 The molar mass of benzene is $78$ g/mol.
${\text{90}}{\,^{\text{o}}}{\text{C}}$ is ${\text{1020}}$ torr. A solution of $5$ g of a solute in $58.5$g benzene has vapour pressure $990$torr. The molecular mass of the solute is:
On substituting ${\text{1020}}$ torr for ${{\text{p}}_{\text{o}}}$, $990$ torr for${{\text{p}}_{\text{1}}}$, $5$ g for ${{\text{w}}_2}$(solute), $58.5$g for${{\text{w}}_1}$( benzene), and $78$ for ${{\text{M}}_1}$( benzene),
$\dfrac{{1020\, - 990}}{{1020}}{\text{ = }}\dfrac{{5 \times 78}}{{{{\text{M}}_2}\, \times 58.5}}$
$\dfrac{{30}}{{1020}}{\text{ = }}\dfrac{{390}}{{{{\text{M}}_2}\, \times 58.5}}$
$0.03{\text{ = }}\dfrac{{6.7}}{{{{\text{M}}_2}}}$
${{\text{M}}_2}{\text{ = }}\dfrac{{6.7}}{{\,0.03}}$
${{\text{M}}_2}{\text{ = }}222$ g/mol
So, the molecular mass of the solute is $222$ g/mol which is near to $220$ g/mol.


Therefore, option (D) is correct.


Note: The w/M, (where, w is the mass and M is the molecular mass), is known as moles. For the two substances, one is solvent and one is solute, the relative lowering of vapour pressure is directly proportional to the number of moles of solute and inversely proportional to the mole of solvent. When we add a non-volatile solute in the solvent, the vapour pressure of the solution decreases. This is known as relative lowering of vapour pressure.