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Question: 0.24g of a volatile liquid upon vaporisation gives \[\text{45 ml}\] of vapours at STP. What will be ...

0.24g of a volatile liquid upon vaporisation gives 45 ml\text{45 ml} of vapours at STP. What will be the vapour density of the substance? (Density of H2= 0.089 gL-1{{\text{H}}_{\text{2}}}\text{= 0}\text{.089 g}{{\text{L}}^{\text{-1}}}).
a.) 9.539
b.) 59.73
c.) 5.993
d.) 95.39

Explanation

Solution

Hint: In order to solve the numerical, we need to have an idea about the concept of vapour density. Vapour density is defined as the ratio between of a vapour in relation to that of hydrogen.
Formula used: Vapour Density = Mass of n number of molecules of gasMass of n number molecules of hydrogen \text{Vapour Density = }\dfrac{\text{Mass of n number of molecules of gas}}{\text{Mass of n number molecules of hydrogen}}\text{ }

= Mass of certain volume of the vapourMass of an equal volume of hydrogen measured under the same conditions of T and P\text{= }\dfrac{\text{Mass of certain volume of the vapour}}{\text{Mass of an equal volume of hydrogen measured under the same conditions of T and P}}

Complete step-by-step answer:
Vapour Density = Mass of n number of molecules of gasMass of n number molecules of hydrogen \text{Vapour Density = }\dfrac{\text{Mass of n number of molecules of gas}}{\text{Mass of n number molecules of hydrogen}}\text{ }
= 0.24g45 !!×!! 10-3 L0.089gL-1 = 59.9\text{= }\dfrac{\dfrac{\text{0}\text{.24g}}{\text{45}}\text{ }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ L}}{\text{0}\text{.089g}{{\text{L}}^{\text{-1}}}\text{ }}\text{= 59}\text{.9}
Alternatively, mass of 45mL of vapour (NTP) = 0.24g\text{= 0}\text{.24g}
Therefore, Mass of 22,400mL of vapour (NTP)
=0.2445 !!×!! 22400g = 119.5g\text{=}\dfrac{\text{0}\text{.24}}{\text{45}}\text{ }\\!\\!\times\\!\\!\text{ 22400g = 119}\text{.5g}
Hence molecular mass = 119.5u\text{= 119}\text{.5u}
We know,
Vapour density = molecular mass2 = 59.75\text{Vapour density = }\dfrac{\text{molecular mass}}{\text{2}}\text{ = 59}\text{.75}
So the answer to the above question can be termed as 59.73, since it is the closest round of answer. So the answer to the question is Option B.

Note: The concept of vapour density is important because it will give an indication where the gas or vapours can generally be expected to be located at hazmat releases.
Knowing the vapour density also helps us in determining the molar mass of the substance. If the substance in question is volatile liquid, a common method to find the molar mass is to vaporize it and apply the ideal gas law, PV=nRT\text{PV=nRT},
where,
P= pressure
V = volume
n = amount of substance
R = ideal gas constant
T = temperature