Question

Find the relative molecular mass of methyl alcohol $\left( {C{H_3}OH} \right)$ , if $160gm$ of the alcohol on vaporization has a volume of $112$ litres at STP.
(A) $16gm$
(B) $32gm$
(C) $160gm$
(D) $80gm$

Answer 1

Here, we use the concept of moles to calculate the amount of water formed. It can be calculated by dividing the mass of a substance present in a mixture by its molar mass:
${\text{The number of moles of a substance }} = \,\dfrac{{{\text{Mass of the substance in the mixture}}}}{{{\text{Molar mass of the substance}}}}$
Another important concept coming into use is that one mole of a gaseous substance occupies $22.4\;litres$volume.

Complete answer:
The questions requires us to calculate the relative molar mass of methyl alcohol under given conditions, where mass $(w)$ and volume of the alcohol vaporized is $160gm$ and $112\,litres$(at STP) respectively.
We know that one mole of a gaseous substance occupies $22.4\;litres$volume. Using this relation, we can calculate the number of moles occupied by $112\,litres$.
${\text{Number of moles occupied by }}112\,litres{\text{ of gas }}\,{\text{ = }}\,\,\dfrac{{112}}{{22.4}} = \,\,5\,moles\,\,\, - - - (1)$
Since we know the mass of methyl alcohol vaporized, we can substitute it in the following equation relating number of moles:
${\text{The number of moles of a substance }} = \,\dfrac{{{\text{Mass of the substance in the mixture}}}}{{{\text{Molar mass of the substance}}}}$
$= \dfrac{w}{M}$
$= \dfrac{{160}}{M}\,\,\,\,\, - - - (2)$
In $(2),$ we can substitute with $(1)$ in the LHS. On rearranging the equation so as to calculate relative molar mass $(M)$of methyl alcohol, we get:
$M\, = \,\dfrac{{160}}{5}$
$= 32gm$
Hence, the relative molecular mass of methyl alcohol $\left( {C{H_3}OH} \right)$ , if $160gm$ of the alcohol on vaporization has a volume of $112$ litres at STP is (B) $32gm$ .

Note:
There exists an equation relating vapour density and molar mass, where:
${\text{Molar}}\,{\text{mass}}\, = \,2 \times \,{\text{Vapour}}\,{\text{density}}$
This equation comes into play if we are provided with the vapour density of the substance and are supposed to find its molar mass, or vice versa.