Question

If the weight of 5.6 litres of a gas at $\mathrm{N.T.P.}$ is 11 grams. The gas may be:
$\mathrm{A}) \mathrm{PH}_{3}$
$\mathrm{B}) \mathrm{CoCl}_{2}$
$\mathrm{C}) \mathrm{NO}$
$\mathrm{D}) \mathrm{N}_{2} \mathrm{O}$

Answer 1

We know that volume occupied by 1 mole of gas = 22.4 litres at N.T.P. Then, we will calculate the molar mass of all the options and find out the compound.

Complete step-by-step answer:
- Putting the value of temperature , Pressure into the ideal gas equation at STP/NTP into the ideal gas equation PV= nRT
i.e T = 273 K
and P = 1atm
V/n = 22.4L/mole
- Hence, we come to know that 1 mole of any gas will contain 22.4 L at NTP.
- Thus, moles of gas = (5.6/22.4) = 0.25.
- Since, 0.25 moles weigh 11 gram, 1 mole will weigh 44 grams.
Therefore, we need a gas whose molar mass is 44 grams.
- Discussing option A, let us calculate the molar mass of phosphine i.e., (31+3 = 34 g/mol)
Thus, it will not have 11 grams at 5.6 L at NTP.
- Discussing option B, let us calculate the molar mass of cobalt chloride i.e., (130 g/mol)
Thus, it will not have 11 grams at 5.6 L at NTP.
- Discussing option C, let us calculate the molar mass of nitric oxide i.e., (30 g/mol)
Thus, it will not have 11 grams at 5.6 L at NTP.
- Discussing option D, let us calculate the molar mass of nitrous oxide i.e., (44 g/mol)
Thus, it will have 11 grams at 5.6 L at NTP.
Clearly, option D is the answer.

Additional information: Nitrous oxide, commonly known as laughing gas or nitrous, is a chemical compound, an oxide of nitrogen with formula $\mathrm{N}_{2} \mathrm{O}$.
At room temperature, it is a colorless non- flammable gas, with a slight metallic scent and taste. At elevated temperatures, nitrous oxide is a powerful oxidiser similar to molecular oxygen. It is soluble in water.

Note: Always remember that 1 mole of gas has 22.4 L at NTP/STP. This is only when the pressure is considered as 1 atm not 1 bar.