
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
510.3k+ views
Hint: 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.
\[{\text{The number of moles of a substance }} = \,\dfrac{{{\text{Mass of the substance in the mixture}}}}{{{\text{Molar mass of the substance}}}}\]
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.
Recently Updated Pages
Give the total number of monochloro products including class 11 chemistry CBSE

Describe the dihybrid cross experiment performed by class 11 biology CBSE

Study the pedigree chart given below What does it show class 11 biology CBSE

In the pulley system shown if radii of the bigger and class 11 physics CBSE

Find the rank of the word MOTHER in dictionary for class 11 maths CBSE

Study the various instances of the transformation of class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

