Question

The V.D of a gas is $11.2$. The volume occupied by $\text{11}\text{.2gm}$ of this gas at N.T.P is:
(A) $\text{22}\text{.4 litres}$
(B) $\text{11}\text{.2 litres}$
(C) $\text{1 litres}$
(D) $\text{2}\text{.24 litres}$

Answer 1

Hint: In this question V.D refers to vapour density. Vapour density of a gas is the ratio of density of gas and density of hydrogen at the same temperature and pressure. $\text{Vapour}\,\text{density}=\dfrac{\text{Density}\,\text{of}\,\text{gas}\,\text{(}{{\text{d}}_{\text{gas}}}\text{)}}{\text{Density}\,\text{of}\,\text{hydrogen}\,\text{(}{{\text{d}}_{{{\text{H}}_{\text{2}}}}}\text{)}}.........(i)$
N.T.P stands for Normal Temperature and Pressure. At N.T.P pressure of gas is 1atm and temperature is 293K. At N.T.P, 1mole of any gaseous substance occupies$\text{22}\text{.4 litres}$of volume, and this volume is called gram molar volume.

Complete step by step solution:
Calculation of volume occupied by gas is calculated in three steps-
In first step we will calculate the molar mass of gas by applying equation (i)
Since $\text{Vapour}\,\text{density}=\dfrac{{{d}_{gas}}}{{{d}_{{{H}_{2}}}}}=\dfrac{{{\text{m}}_{\text{gas}}}\,\text{for certain}\,\text{V}\,\text{litre}\,\text{volume}}{{{\text{m}}_{{{\text{H}}_{\text{2}}}}}\text{for certain}\,\text{V}\,\text{litre}\,\text{volume}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\{\,\text{d=}\dfrac{\text{m (mass)}}{\text{V(volume)}}\}$
So if N molecule present in the given volume of a gas and hydrogen at N.T.P
$\text{V}\text{.D =}\dfrac{{{\text{m}}_{\text{gas}}}\text{of}\,\text{N}\,\text{molecule}}{{{\text{m}}_{{{\text{H}}_{\text{2}}}}}\text{of}\,\text{N molecule}}\,=\,\dfrac{{{\text{m}}_{\text{gas}}}\text{of}\,1\,\text{molecule}}{{{\text{m}}_{{{\text{H}}_{\text{2}}}}}\text{of}\,1\text{ molecule}}\,=\dfrac{\text{molecular}\,\text{mass}\,\text{of}\,\text{gas}}{\text{2}}..........(ii)$
By applying equation (ii) we will calculate the molecular mass of gas after putting V.D = 11.2
\[\begin{align}
& \text{V}\text{.D =}\,\dfrac{\text{molar}\,\text{mass}\,\text{of}\,\text{gas}}{\text{2}} \\
& \text{molar}\,\text{mass}\,=\,\,2\,\times \,\,\text{V}\text{.D} \\
& \text{molar}\,\text{mass}\,=\,\,2\,\times \,\,11.2 \\
& \text{molar}\,\text{mass}\,=\,\,22.\text{4gm} \\
\end{align}\]
In second step we will calculate the number of mole present in given mass of gas.
So number of moles present in 11.2gm of gas
\[\begin{align}
& \text{mole}\,\text{(n) = }\dfrac{\text{given}\,\,\text{mass}}{\text{molar}\,\text{mass}} \\
& \text{n}=\dfrac{11.2}{22.4} \\
&\therefore \text{n}=\dfrac{1}{2} \\
\end{align}\]
In the third step we will calculate the occupied volume of gas by applying a unitary method.
Since at N.T.P 1mole of any gaseous substance occupies $\text{22}\text{.4 litres}$ of volume
If one mole of given gas occupies $\text{22}\text{.4 litres}$, so half mole of gas will occupy $\text{11}\text{.2 litres}$ of gas.

So option (B) will be the correct answer.

Note: Vapour density is calculated with respect to the hydrogen gas under similar conditions of temperature and pressure.
Vapour density, relative density and specific gravity are the ratios, so these are unit less quantities.