Question

$896mL$ of a mixture of $CO$ and $C{O_2}$ weigh $1.28g$ at NTP. Calculate the volume of $C{O_2}$ in the mixture at NTP.

Answer 1

Volume of one mole of a gas at NTP is $22400mL$. At NTP volume of one mole of any gas is equal to $22400mL$. Number of particles in one mole of a substance is constant and this number is equal to $6.02 \times {10^{23}}$. One mole of any substance will contain $6.02 \times {10^{23}}$ number of molecules or atoms.

Formula used: Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$
Mole fraction of a substance$= \dfrac{{{\text{number of moles of that substance}}}}{{{\text{total number of moles}}}}$
Volume of one mole of gas at NTP $= 22400mL$

Complete step by step answer:
In this question we have been given a mixture of $CO$ and $C{O_2}$ at NTP. Mass of the mixture is $1.28g$ and volume of the mixture is $896mL$. We have to find the volume of $C{O_2}$ at NTP.
For this we have to find the mass of each gas. Let mass of $C{O_2}$ be $x$. Total mass of the mixture is $1.28g$ so, if mass of $C{O_2}$ is $x$, then mass of $CO$ will be $1.28 - x$.
Molecular mass is the mass of one mole of a substance. One mole of any substance contains $6.02 \times {10^{23}}$ number of molecules or atoms. Molecular mass can be calculated by adding mass of individual atoms of a molecule.
Molecular mass of $CO$ is $28g$ (mass of carbon atom is $12g$ and mass of oxygen atom is $16g$).
Molecular mass of $C{O_2}$ is $44g$ (mass of carbon atom is $12g$ and mass of oxygen atom is $16g$ but in this molecule there are two atoms of oxygen).
From this mass (calculated above) and molecular mass we can calculate number of moles $CO$ and $C{O_2}$. Formula to calculate number of moles is:
Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$
For $CO$, mass given is (calculated above) $1.28 - x$ and molecular mass is $28g$. Therefore,
Number of moles of $CO = \dfrac{{1.28 - x}}{{28}}$
For $C{O_2}$, mass given is $x$ (calculated above) and molecular mass is $44g$. Therefore,
Number of moles of $C{O_2} = \dfrac{x}{{44}}$
We know that volume of one mole of gas at NTP is $22400mL$. Therefore,
$22400mL{\text{ of gas }} = 1mole$
Or we can say that,
$1mL{\text{ of gas }} = \dfrac{1}{{22400}}mole$
Given volume of gas is $896mL$. Therefore,
$896mL{\text{ of gas }} = \dfrac{1}{{22400}} \times 896mole$
Solving this we get,
Number of moles$= 0.04$
Number of moles of mixture of gas is equal to the total moles. So, total moles is equal to $0.04$
From the number of moles of mixture, $CO$ and $C{O_2}$ we can calculate mole fraction of $CO$ and $C{O_2}$. Formula to calculate mole fraction is:
Mole fraction of a substance$= \dfrac{{{\text{number of moles of that substance}}}}{{{\text{total number of moles}}}}$
Mole fraction of $CO$ $= \dfrac{{\dfrac{{1.28 - x}}{{28}}}}{{0.04}} = \dfrac{{1.28 - x}}{{1.12}}$ (number of moles of $CO$ and mixture is calculated above)
Similarly,
Mole fraction of $C{O_2}$ $= \dfrac{{\dfrac{x}{{44}}}}{{0.04}} = \dfrac{x}{{1.76}}$ (number of moles of $C{O_2}$ and mixture is calculated above)
Sum of mole fraction of individual components of a mixture is equal to one. This mixture is made of $CO$ and $C{O_2}$. Therefore the sum of mole fraction of $CO$ and $C{O_2}$ is one.
So,
$\dfrac{{1.28 - x}}{{1.12}} + \dfrac{x}{{1.76}} = 1$ (Substituting value of mole fraction of components which is calculated above)
Solving this we get, $x = 0.44g$
Therefore given mass of $C{O_2}$ is $0.44g$. Molecular mass of $C{O_2}$ is $44g$. So,
Number of moles of $C{O_2} = \dfrac{{0.44}}{{44}} = 0.01$
We know that the volume of one mole of gas at NTP is $22400mL$. Therefore,
$1mole = 22400mL{\text{ of gas }}$
Moles of $C{O_2}$ are $0.01$. So,
$0.01mole = \left( {22400 \times 0.01} \right)mL{\text{ of gas }}$
Solving this we get a volume of $C{O_2}$ equal to $224mL$.

Note:
Sum of mole fraction of individual components of a mixture is equal to one.
Number of particles in one mole of a substance doesn’t depend upon temperature and pressure. This number is the same for all the substances and is equal to $6.02 \times {10^{23}}$.