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Question: The gaseous mixture of \({H_2}\) and \(C{O_2}\)contains \(66\) mass \(\% \) of \(C{O_2}\). The vapou...

The gaseous mixture of H2{H_2} and CO2C{O_2}contains 6666 mass %\% of CO2C{O_2}. The vapour density of the mixture is:
a) 6.16.1
b) 5.45.4
c) 2.72.7
d) 10.810.8

Explanation

Solution

To solve such questions,we should have Knowledge about formula of mass %, vapour density average mass is necessary. The relation of vapour density with average mass is necessary which is mentioned in the solution.

Complete step by step solution:
Mass%, molarity, molality, normality are the different concentration terms used to define the concentration of a compound in some medium.
Vapour density is a different kind of density used for a particular gas or vapour relative to hydrogen gas at the same temperature and pressure.
vapourdensity(ρ)=mavgmolarmassofH2vapour \,density(\rho ) = \dfrac{{{m_{avg}}}}{{molar\, mass\, of\,{H_2}}}
In this question, mass% of CO2C{O_2} i.e. 66%66\% is given.
Mass% is used here to determine the given mass of CO2C{O_2} in total mass of solution. The given mass of CO2C{O_2} is 66gm66 gm . In the solution hydrogen gas ( H2{H_2} ) is also present. As the total mass of solution is 100100, the given mass of hydrogen ( H2{H_2}) will be 10066=34gm100 - 66 = 34 gm .
Molar mass of CO2=12+2×16=44gmC{O_2} = 12 + 2 \times 16 = 44gm and the molar mass of H2=2×1=2gm{H_2} = 2 \times 1 = 2 gm
Now, the calculation of moles of CO2=givenmassmolarmassC{O_2} = \dfrac{{given\, mass}}{{molar\, mass}}
CO2=6644 CO2=1.5moles  \Rightarrow C{O_2} = \dfrac{{66}}{{44}} \\\ \Rightarrow C{O_2} = 1.5moles \\\
And the moles of hydrogen gas;H2=givenmassmolarmass{H_2} = \dfrac{{given\, mass}}{{molar\, mass}}
H2=342 H2=17moles  \Rightarrow {H_2} = \dfrac{{34}}{2} \\\ \Rightarrow {H_2} = 17moles \\\
Hence the total moles is equal to molesCO2+molesofH2moles \,C{O_2} + moles\, of\,{H_2}
1.5+17=18.5moles\Rightarrow 1.5 + 17 = 18.5moles
Substituting the value of above equation;
As the mavg=totalmassofsolutiontotalmoles{m_{avg}} = \dfrac{{total\, mass \,of \,solution}}{{total\, moles}}
mavg=10018.5=5.4\Rightarrow {m_{avg}} = \dfrac{{100}}{{18.5}} = 5.4
Now to calculate vapour density, the formula is ρ=mavg2\rho = \dfrac{{{m_{avg}}}}{2} , by substituting the values of above equation again in this formula, we get
ρ=5.42 ρ=2.7  \rho = \dfrac{{5.4}}{2} \\\ \Rightarrow \rho = 2.7 \\\
Hence the value of vapour density is equal to 2.72.7
Hence,The correct option is (c).

Note: In chemistry laboratories we use the different concentration terms to define concentration of compound. But in practical use molality is considered as the best option as compared to molarity because volume of a solution can change with temperature but mass of solvent remains constant.