Question

A flask of gaseous $CC{l_4}$ was weighed at measured temperature and pressure. The flask was then flushed and filled with ${O_2}$ at the same temperature and pressure. The weight of $CC{l_4}$ vapour will be-
A. Five times as heavy as ${O_2}$
B. One fifth heavy as compared to ${O_2}$
C. Same as that of ${O_2}$
D. Twice as heavy as ${O_2}$

Answer 1

Hint: First use the ideal gas equation
$ \Rightarrow \dfrac{{{n_1}{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{n_2}{P_2}{V_2}}}{{{T_2}}}$ Where ${P_1}{\text{ and }}{P_2}$ are the pressures of $CC{l_4}$ and ${O_2}$ gas respectively
And ${V_1}{\text{ and }}{V_2}$ are the volumes of $CC{l_4}$ and ${O_2}$ gas respectively. Also ${T_1}$ and ${T_2}$ are temperatures of the gases respectively. And ${n_1}$ and ${n_2}$ are moles of $CC{l_4}$and ${O_2}$gas respectively. Then use the formula of moles to find the ratio between the weight of both gases which is given as-
$ \Rightarrow n = \dfrac{w}{m}$
Where w is the weight and m is the molecular weight of the given gas

Step-by-Step Solution-
Given, a flask of gaseous $CC{l_4}$ was weighed at measured temperature and pressure so let the volume of the flask be V. Now, the flask was then flushed and filled with ${O_2}$ at the same temperature and pressure. So let the temperature be T and pressure be P.
Now we know that according to ideal gas equation-
$ \Rightarrow \dfrac{{{n_1}{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{n_2}{P_2}{V_2}}}{{{T_2}}}$
Where ${P_1}{\text{ and }}{P_2}$ are the pressures of $CC{l_4}$ and ${O_2}$ gas respectively
And ${V_1}{\text{ and }}{V_2}$ are the volumes of $CC{l_4}$ and ${O_2}$ gas respectively. Also ${T_1}$ and ${T_2}$ are temperatures of the gases respectively. And ${n_1}$ and ${n_2}$ are moles of $CC{l_4}$ and ${O_2}$ gas respectively
Since the pressure and temperature are same and the volume of the flask is the same no matter which gas is filled on the flask, then on putting the given values in the equation we get,
$ \Rightarrow $ \[\dfrac{{{n_1}PV}}{T} = \dfrac{{{n_2}PV}}{T}\]
So we get,
$ \Rightarrow {n_1} = {n_2}$ -- (i)
This means that the numbers of moles of the gases are also equal.
Now we know that number of moles is given by the formula-
$ \Rightarrow n = \dfrac{w}{m}$
Where w is the weight and m is the molecular weight of the given gas.
We know that
Molecular weight of $CC{l_4}$=$154g$
And molecular weight of ${O_2}$=$32g$
Then ${n_1} = \dfrac{{{w_{CC{l_4}}}}}{{154}}$ -- (ii)
And ${n_2} = \dfrac{{{w_{{O_2}}}}}{{32}}$ --- (iii)
On substituting the value of eq. (ii) and (iii) in eq. (i) we get,
$ \Rightarrow \dfrac{{{w_{CC{l_4}}}}}{{154}} = \dfrac{{{w_{{O_2}}}}}{{32}}$
Then on solving we get,
$ \Rightarrow \dfrac{{{w_{CC{l_4}}}}}{{{w_{{O_2}}}}} = \dfrac{{154}}{{32}}$
On simplifying we get,
$ \Rightarrow \dfrac{{{w_{CC{l_4}}}}}{{{w_{{O_2}}}}} = 4.8125 \simeq 5$
So the weight of $CC{l_4}$ vapor is five times as heavy as ${O_2}$

Answer- Hence the correct answer is A.

Note: $CC{l_4}$ is known as carbon tetrachloride. It is a clear colorless which is present as gas in the air liquid. It has sweet ether like odour. Its uses are-
- It is used as a solvent and in preparation of other chemicals.
- It is used as a dry cleaning agent, fire extinguisher and grain fumigant.