Question

If the weight of 5.6 litres of a gas at NTP is 11 gram. The gas may be:
(A) $P{H_3}$
(B) $COC{l_2}$
(C) NO
(D) ${N_2}O$

Answer 1

Hint: NTP stands for normal temperature and pressure. At NTP, any gas with 1 mole quantity has the volume of 22.4 liter. We can find the number of moles by dividing the weight of the substance by its molecular weight.

Complete step by step solution:
We are provided the volume of the gas at NTP and we need to find the gas from the given options. So, we will find the moles of the gas and find the gas from it.
- NTP stands for normal temperature and pressure. It indicates that the pressure of the gas is 1 atm and the volume of the gas is 5.6 litres.
- Now, any gas that has a volume of 22.4 Litre, has a quantity of 1 mole. So, we can say that 5.6 litres of the gas would have $\dfrac{{5.6 \times 1}}{{22.4}} = 0.25 mol$ quantity.
- So, we can say that the amount of gas present here is 0.25 mol.
- We know that the number of moles of a substance can be obtained by dividing its weight by its molecular weight.
So, we can write that the number of moles of the gas = $\dfrac{{{\text{Weight}}}}{{{\text{Molecular weight}}}}$ ………(1)
We know that the number of moles of gas is 0.25moles and the weight of gas is given that it is 11g. So, we can write equation (1) as
$0.25 = \dfrac{{11}}{{M.W.}}$
So, we can write that
$M.W. = \dfrac{{11}}{{0.25}} = 44 g$
So, if we calculate the molecular weights of all the given compounds, we obtain that M.W. of ${N_2}O$ is 44$ gmmo{l^{ - 1}}$.
- M.W. of $P{H_3},COC{l_2}$ and NO is 34, 98 and 28$ gmmo{l^{ - 1}}$ respectively.
Thus, we can say that the gas is ${N_2}O$.

So, the correct answer is (D).

Note: Remember that STP stands for standard temperature and pressure and it indicates the temperature is ${0^ \circ }C$ and the pressure is 1 bar. Note that any 1 mole gas has a volume of 22.4 liter at STP.