Question

The gaseous mixture of ${H_2}$ and $C{O_2}$contains $66$ mass $\% $ of $C{O_2}$. The vapour density of the mixture is:
a) $6.1$
b) $5.4$
c) $2.7$
d) $10.8$

Answer 1

Hint: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.
$vapour \,density(\rho ) = \dfrac{{{m_{avg}}}}{{molar\, mass\, of\,{H_2}}}$
In this question, mass% of $C{O_2}$ i.e. $66\% $ is given.
Mass% is used here to determine the given mass of $C{O_2}$ in total mass of solution. The given mass of $C{O_2}$ is $66 gm$ . In the solution hydrogen gas ( ${H_2}$ ) is also present. As the total mass of solution is $100$, the given mass of hydrogen ( ${H_2}$) will be $100 - 66 = 34 gm$ .
Molar mass of $C{O_2} = 12 + 2 \times 16 = 44gm$ and the molar mass of ${H_2} = 2 \times 1 = 2 gm$
Now, the calculation of moles of $C{O_2} = \dfrac{{given\, mass}}{{molar\, mass}}$
  $
\Rightarrow C{O_2} = \dfrac{{66}}{{44}} \\
\Rightarrow C{O_2} = 1.5moles \\
$
And the moles of hydrogen gas;${H_2} = \dfrac{{given\, mass}}{{molar\, mass}}$
$
\Rightarrow {H_2} = \dfrac{{34}}{2} \\
\Rightarrow {H_2} = 17moles \\
$
Hence the total moles is equal to $moles \,C{O_2} + moles\, of\,{H_2}$
$ \Rightarrow 1.5 + 17 = 18.5moles$
Substituting the value of above equation;
As the ${m_{avg}} = \dfrac{{total\, mass \,of \,solution}}{{total\, moles}}$
$ \Rightarrow {m_{avg}} = \dfrac{{100}}{{18.5}} = 5.4$
Now to calculate vapour density, the formula is $\rho = \dfrac{{{m_{avg}}}}{2}$ , by substituting the values of above equation again in this formula, we get
$
\rho = \dfrac{{5.4}}{2} \\
\Rightarrow \rho = 2.7 \\
$
Hence the value of vapour density is equal to $2.7$
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.