
Taking the vapour density of hydrogen gas as 1. What will be the vapour density of $C{{H}_{4}}$?
Answer
566.4k+ views
Hint: Think about the relation between molecular mass and vapour density and the definition of vapour density. The fact that molecular mass is usually calculated relative to the mass of 1 atom of hydrogen should also be taken into consideration.
Vapour density is a new concept and is defined as the relative density of vapour or any particular gas with respect to the density of hydrogen gas at the same temperature and pressure.
Complete answer:
Vapour density is defined as the relative density of vapour or any particular gas with respect to the density of hydrogen gas at the same temperature and pressure. Hence, the formula of vapour density is noted as:
\[\text{Vapour density =}\dfrac{\text{mass of a certain volume of gas or vapour}}{\text{mass of the same volume of hydrogen gas}}\]
Let us consider that the predefined volume mentioned here is the volume occupied by one molecule of the respective elements. Hence, we can modify the formula into:
\[\text{Vapour density = }\dfrac{\text{mass of one molecule of a certain gas of vapour}}{\text{mass of one molecule of hydrogen}}\]
We already know the formula used to calculate the molecular mass of any molecule.
\[\text{Molecular mass = }\dfrac{\text{mass of one molecule of a substance}}{\text{mass of one atom of hydrogen}}\]
Now, to relate the molecular mass and vapour density, we know that hydrogen is a diatomic gas. Hence, the mass of one molecule of hydrogen will be twice as that of one atom of hydrogen. Thus, the formula for vapour density can be modified to:
\[\text{Vapour density = }\dfrac{\text{mass of one molecule of a certain gas of vapour}}{\text{2}\times \text{mass of one atom of hydrogen}}\]
Now, putting the equation for molecular mass into the one for vapour density, we get:
\[\text{Vapour density = }\dfrac{\text{molecular mass}}{2}\]
The molecular mass of $C{{H}_{4}}$is $(1\times \text{mass of }C)+(4\times \text{mass of }H)$
Molecular mass of $C{{H}_{4}}$= $(1\times 12)+(4\times 1)$
Molecular mass of $C{{H}_{4}}$= 16
Therefore,
\[\text{Vapour density of }C{{H}_{4}}=\dfrac{16}{2}\]
Vapour density of $C{{H}_{4}}$ is 8
Note:
Please be careful while handling whether the molecular mass of hydrogen is to be considered or the atomic mass. Consider atomic mass for molecular weight and consider molecular mass for vapour density as in vapour form hydrogen is found in the ${{H}_{2}}$ state.
Vapour density is a new concept and is defined as the relative density of vapour or any particular gas with respect to the density of hydrogen gas at the same temperature and pressure.
Complete answer:
Vapour density is defined as the relative density of vapour or any particular gas with respect to the density of hydrogen gas at the same temperature and pressure. Hence, the formula of vapour density is noted as:
\[\text{Vapour density =}\dfrac{\text{mass of a certain volume of gas or vapour}}{\text{mass of the same volume of hydrogen gas}}\]
Let us consider that the predefined volume mentioned here is the volume occupied by one molecule of the respective elements. Hence, we can modify the formula into:
\[\text{Vapour density = }\dfrac{\text{mass of one molecule of a certain gas of vapour}}{\text{mass of one molecule of hydrogen}}\]
We already know the formula used to calculate the molecular mass of any molecule.
\[\text{Molecular mass = }\dfrac{\text{mass of one molecule of a substance}}{\text{mass of one atom of hydrogen}}\]
Now, to relate the molecular mass and vapour density, we know that hydrogen is a diatomic gas. Hence, the mass of one molecule of hydrogen will be twice as that of one atom of hydrogen. Thus, the formula for vapour density can be modified to:
\[\text{Vapour density = }\dfrac{\text{mass of one molecule of a certain gas of vapour}}{\text{2}\times \text{mass of one atom of hydrogen}}\]
Now, putting the equation for molecular mass into the one for vapour density, we get:
\[\text{Vapour density = }\dfrac{\text{molecular mass}}{2}\]
The molecular mass of $C{{H}_{4}}$is $(1\times \text{mass of }C)+(4\times \text{mass of }H)$
Molecular mass of $C{{H}_{4}}$= $(1\times 12)+(4\times 1)$
Molecular mass of $C{{H}_{4}}$= 16
Therefore,
\[\text{Vapour density of }C{{H}_{4}}=\dfrac{16}{2}\]
Vapour density of $C{{H}_{4}}$ is 8
Note:
Please be careful while handling whether the molecular mass of hydrogen is to be considered or the atomic mass. Consider atomic mass for molecular weight and consider molecular mass for vapour density as in vapour form hydrogen is found in the ${{H}_{2}}$ state.
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