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A gas cylinder is filled with hydrogen gas which weighs \[40\text{ }g\]. The same cylinder holds \[\text{880 g}\] of a gas A and \[\text{560 g}\] of a gas B under the same conditions of temperature and pressure. Calculate the relative molecular masses of A and B.
(A) $\text{A}\to \text{44 , B}\to \text{28}$
(B) $\text{A}\to 32\text{ , B}\to 64$
(C) $\text{A}\to \text{46 , B}\to \text{28}$
(D) $\text{A}\to \text{44 , B}\to 32$

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Last updated date: 13th Jun 2024
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Answer
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Hint: The container holds the hydrogen gas, gas A, and gas B at the same pressure and temperature. Since the gas is held in the same container thus volume would be the same. If the pressure, temperature, and volume remain constant the number of moles for all the gases would be equal. We can now calculate the molar mass of the gas A and B.

Complete step by step solution:
For the ideal gas equation establishes the relation between the pressure, absolute temperature, gas constant, number of moles, and volume as shown below,
$\text{PV = nRT}$
We know that the number of moles is the ratio of mass to the molar mass. We get,
$\text{n = }\dfrac{\text{W}}{\text{M}}$
Let’s substitute the values we get,
$\begin{align}
& \text{PV = }\left( \dfrac{\text{W}}{\text{M}} \right)\text{ R T} \\
& \Rightarrow \dfrac{\text{W}}{\text{M}}=\text{ }\dfrac{\text{P V}}{\text{R T}} \\
\end{align}$
The volume of the container, temperature, and pressure are the same for all the gases.
Therefore, the number of moles is a constant term.
$\dfrac{\text{W}}{\text{M}}\text{ = }\dfrac{\text{P V}}{\text{R T}}=\text{Constant}$
The hydrogen gas is filled in the container and the x moles of hydrogen gas are present. If the same container is filled with gas A and gas B then both will contain the same number of moles.
The number of moles of the gas A and gas B will be equal to the moles of the hydrogen gas.
\[\begin{matrix}
\text{No}\text{.of moles of }{{\text{H}}_{\text{2}}}\text{ gas} & = & \text{No}\text{.of moles of A gas} & = & \text{No}\text{.of moles of B gas} \\
\dfrac{{{\text{W}}_{{{\text{H}}_{\text{2}}}}}}{{{\text{M}}_{{{\text{H}}_{\text{2}}}}}} & = & \dfrac{{{\text{W}}_{\text{A}}}}{{{\text{M}}_{\text{A}}}} & = & \dfrac{{{\text{W}}_{\text{B}}}}{{{\text{M}}_{\text{B}}}} \\
\dfrac{40\text{ }}{2\text{ g mo}{{\text{l}}^{\text{-1}}}} & = & \dfrac{880\text{ g}}{{{\text{M}}_{\text{A}}}} & = & \dfrac{560\text{ g}}{{{\text{M}}_{\text{B}}}} \\
\end{matrix}\]
Let’s simply for the molar mass of A, we get
\[\begin{align}
& {{\text{M}}_{\text{A}}}\text{= 880 g }\!\!\times\!\!\text{ }\dfrac{\text{2 g mo}{{\text{l}}^{\text{-1}}}}{\text{40 g}} \\
 & \text{ = }\dfrac{\text{1760}}{\text{40}}\text{ g mo}{{\text{l}}^{\text{-1}}} \\
 & \text{ }{{\text{M}}_{\text{A}}}\text{= 44 g mo}{{\text{l}}^{\text{-1}}} \\
\end{align}\]
Let’s calculate the molecular weight of the B.
\[\begin{align}
& {{\text{M}}_{\text{B}}}\text{= 560 g }\!\!\times\!\!\text{ }\dfrac{\text{2 g mo}{{\text{l}}^{\text{-1}}}}{\text{40 g}} \\
 & \text{ = }\dfrac{\text{1120}}{\text{40}}\text{ g mo}{{\text{l}}^{\text{-1}}} \\
 & \text{ }{{\text{M}}_{\text{B}}}\text{= 28 g mo}{{\text{l}}^{\text{-1}}} \\
\end{align}\]
Therefore, the molecular weight of A gas is \[\text{44 g mo}{{\text{l}}^{\text{-1}}}\] and the molecular weight of B gas is \[\text{28 g mo}{{\text{l}}^{\text{-1}}}\].
Hence the relative molar mass are, $\text{A}\to \text{44 , B}\to \text{28}$

Hence, (A) is the correct option.

Note: We are considering the same container throughout the problem. Therefore, the volume of the gas occupied will always be the same. Since we are considering the same temperature, pressure on the container, the number of moles would be the same too. This is also at the STP condition. This is standard temperature pressure, where pressure is $\text{1}{{\text{0}}^{\text{5}}}\text{ Pa}$ and temperature of $\text{273}\text{.15 K}$.