Question

Equal weight of $CO$ and $C{H_4}$ are mixed together in an empty container at $300K$ . The fraction of total pressure exerted by $C{H_4}$ is:
A) $\dfrac{{16}}{{17}}$
B) $\dfrac{7}{{11}}$
C) $\dfrac{8}{9}$
D) $\dfrac{5}{{16}}$

Answer 1

To solve this question first we need to find the moles of $CO$ and $C{H_4}$ . Also, we know that the number of moles is equal to weight divided by molecular weight. Since the weight of $CO$ and $C{H_4}$ are unknown in this question so we will take the weight as $(a)grams$ since it is given in the question that both the compounds have equal weight.

Complete step by step answer:
As we know that equal weight of $CO$ and $C{H_4}$ are mixed together in an empty container so,
Let the weight of $CO$ $=$ the weight of $C{H_4}$ $= a$ gm.
So, to calculate the number of moles in $CO$ and $C{H_4}$ we will use the formula:
Number of moles $= \dfrac{{weight}}{{Mol.weight}}$
Number of moles of $CO$ $= \dfrac{a}{{12 + 16}} = \dfrac{a}{{28}}$
Number of moles of $C{H_4}$ $= \dfrac{a}{{12 + 1(4)}} = \dfrac{a}{{16}}$
So, the total number of moles present in the mixture $= \dfrac{a}{{28}} + \dfrac{a}{{16}}$
Now we will find the mole fraction of $C{H_4}$ by using the formula $\dfrac{{Moles(gas)}}{{Moles(total)}}$ as we know mole fraction of a particular gas is equal to the number of moles of the particular gas divided by total number of moles present in the mixture.
So, $xC{H_4} = \dfrac{{\dfrac{a}{{16}}}}{{\dfrac{a}{{28}} + \dfrac{a}{{16}}}}$
$= \dfrac{a}{{16}} \times \dfrac{{28 \times 16}}{{44a}} \\\ = \dfrac{{14}}{{22}} \\\ = \dfrac{7}{{11}} \\\$
We know that as per raoult’s law which states that the pressure exerted by the individual gas is directly proportional to its mole fraction
Which therefore means that the fraction of pressure exerted by the $C{H_4} = \dfrac{7}{{11}}$
Hence the correct answer will be Option B.

Note: For calculating the number of moles of a gas it is important for you to remember the atomic weight or the atomic number of each element in order to find the molecular mass of the compound because most of the times the molecular mass is not mentioned in the question. Also, you should always remember the formula for calculating the number of moles.