Answer

Verified

67.8k+ views

**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.

Recently Updated Pages

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

A man stands at a distance of 250m from a wall He shoots class 9 physics JEE_Main

Other Pages

Electric field due to uniformly charged sphere class 12 physics JEE_Main

A particle is given an initial speed u inside a smooth class 11 physics JEE_Main

A train starts from rest from a station with acceleration class 11 physics JEE_Main

Mulliken scale of electronegativity uses the concept class 11 chemistry JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main