Question

The vapour pressure of pure $A$ is 10 torr and at the same temperature, when $1g$ of $B$ is dissolved in $20g$ of $A$, its vapour pressure is reduced to 9 torr. If the molecular mass of $A$ is 200amu then the molecular mass of $B$ is:
A. 100amu
B. 90amu
C. 75amu
D. 120amu

Answer 1

Consider what is happening after the addition of the solute and recall the formula that is used for the relative lowering of vapour pressure. Convert the number of moles to the weight divided by the molecular weight and then solve for the molecular weight of the solute.

Complete step by step solution:
In this problem, we can see that the vapour pressure is getting lowered when, compared to the vapour pressure of a pure substance, an impurity is added to it. This impurity will be our solute. So, we will use the formula for the relative lowering of vapour pressure to calculate the molecular mass of the solute.
The formula for the relative lowering of vapour pressure states that the reduced vapour pressure caused by the solute divided by the vapour pressure of the pure substance is equal to the number of moles of the solute divided by the total number of moles. The formula is as follows:
$\dfrac{{{P}_{o}}-P}{{{P}_{o}}}=\dfrac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$
Where,
${{P}_{o}}$ = vapour pressure of pure substance
$P$ = reduced vapour pressure after addition of solvent
${{n}_{1}}$ = number of moles of solute
${{n}_{2}}$ = number of moles of solvent
We know that the formula for the number of moles is the weight given divided by the molecular weight of that substance. So, the formula is:
$n=\dfrac{w}{m}$
Where,
$w$ = weight of substance
$m$ = molecular weight of substance
From the given question, we know that:
${{P}_{o}}$ = 10 torr
$P$ = 9 torr
${{w}_{1}}$ = 1 gram
${{w}_{2}}$ = 20 grams
${{m}_{2}}$ = 200 amu
We have modified the equation to:

& \dfrac{{{P}_{o}}-P}{{{P}_{o}}}=\dfrac{\dfrac{{{w}_{1}}}{{{m}_{1}}}}{\dfrac{{{w}_{1}}}{{{m}_{1}}}+\dfrac{{{w}_{2}}}{{{m}_{2}}}} \\\ & \dfrac{{{P}_{o}}-P}{{{P}_{o}}}=\dfrac{{{w}_{1}}{{m}_{2}}}{{{w}_{1}}{{m}_{2}}+{{w}_{2}}{{m}_{1}}} \\\ \end{aligned}$$ Now, we will put the values in the equation. $$\dfrac{10-9}{10}=\dfrac{1\times 200}{(1\times 200)+20\times {{m}_{1}}}$$ Solving for ${{m}_{1}}$ we get: $$\begin{aligned} & 200+20{{m}_{1}}=2000 \\\ & 20{{m}_{1}}=1800 \\\ & {{m}_{1}}=90 \\\ \end{aligned}$$ So, the molecular weight of substance B is 90 amu. Hence, the correct answer is ‘B. 90amu’ **Note:** There is no need to convert all the units to the SI units or some standard units as they are all going to cancel out anyway in the end. Remember to consider the formula for how to calculate the moles of any substance.