Question

A 10 L box contains 41.4g of a mixture of gases ${C_x}{H_8}$ and ${C_x}{H_{12}}$ . The total pressure at ${44^0}C$ in flask is 1.56 atm. Analysis revealed that the gas mixture has 87% total $C$ and 13% total $H$. Find out the value of $x$ .
A) 1
B) 3
C) 5
D) 2

Answer 1

Hint: To find the value of $x$ , first we need to find the total mass of carbon in the mixture in the form of $x$ and then find the observed percentage of carbon in the mixture. Equate this observed percentage with the given percentage of total carbon in the mixture.

Complete step by step answer:
Given data: Pressure (P) = 1.56 atm
Volume (V) = 10 L
Temperature (T) = ${44^0}C$=317 K
Gas constant (R) = $0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$
Using Ideal gas equation:
$PV = nRT$
$n = \dfrac{{PV}}{{RT}}$ --- (1)
where, P= pressure , V= volume, T= temperature, R= gas constant and n= no. of moles of the gas.
Putting the values of given data in equation (1),

We get , $n = \dfrac{{1.56{\text{ atm}} \times 10{\text{ L}}}}{{0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}} \times 317{\text{ K}}}} = 0.6{\text{ mol}}$
$ \Rightarrow $ 0.6 mol are the total moles of the mixture of two given gases.

Let, moles of ${C_x}{H_8}$ be ‘$a$’ , then moles of ${C_x}{H_{12}}$ will be $'0.6{\text{ }} - {\text{ }}a'$ .
Mass of carbon in ‘$a$’ mol of ${C_x}{H_8}$= $12ax{\text{ g}}$
Mass of carbon in $'0.6{\text{ }} - {\text{ }}a'$mol of ${C_x}{H_{12}}$= $12 \times (0.6 - a)x{\text{ g}}$
$\therefore $ Total mass of carbon in the mixture =$12ax + 12(0.6 - a)x = 7.2x{\text{ g}}$
Also, given that total mass of the mixture is 41.4 g .
$\therefore $ percentage of carbon in the mixture = $\dfrac{{{\text{mass of carbon in the mixture }}}}{{{\text{ total mass of the mixture}}}} \times 100$
$ \Rightarrow $ percentage of carbon in the mixture= $\dfrac{{7.2x}}{{41.4}} \times 100$

It is given in the question that the gas mixture contains 87% carbon. So equating the above observed percentage of carbon in the mixture with the given percentage .
$ \Rightarrow $$\dfrac{{7.2x}}{{41.4}} \times 100 = 87$
$\therefore $$x = 5$
So, the correct answer is “Option C”.

Note: Take care of the units of R and its value for different units.
 $\begin{gathered}
  R = 0.08314{\text{ bar d}}{{\text{m}}^3}{\text{mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}
  R = 0.0821{\text{ atm L mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}{\text{ }}
  R{\text{ = 8}}{\text{.314 J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}{\text{ }}
\end{gathered} $
Also, note that mass of a substance = no. of moles $ \times $ atomic/molecular mass.