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

Last updated date: 25th Jun 2024
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Hint: NTP stands for normal temperature and pressure. At normal temperature and pressure, any gas with one mole has a volume of $22.4$ litres. The number of moles is given mass divided by the molecular mass.

We have been given the volume of gas at normal temperature and pressure (NTP). It says that the pressure of the gas is one atm. We will find the number of moles of gas and then the molecular weight says which gas satisfies the given conditions.
Now any gas that has a volume of $22.4$ litres has a quantity of one mole. So it can be concluded that $5.6$ litres of the gas would have $\dfrac{{5.6 \times 1}}{{22.4}} = 0.25$ moles.
So the amount of gas present will be $0.25$ moles. Further, we know that the number of moles of the gas is equal to the given weight of the gas divided by the molecular weight of the gas.
So the number of moles $= \dfrac{m}{W}$
Here the number of moles is $0.25$ and the weight of the gas given is $11$ gram. Putting these values in the given equation we get as follows
$0.25 = \dfrac{{11}}{W}$
$W = \dfrac{{11}}{{0.25}} = 44$ gram
When we calculate the molecular mass of nitrous oxide ${N_2}O$ the molecular weight comes out to be $44\;gmo{l^{ - 1}}$
Here molecular weights of $P{H_3},COC{l_2},NO$ are $34,98,28\;gmo{l^{ - 1}}$ respectively.
Hence the gas is nitrous oxide( ${N_2}O$)

So, the correct answer is Option D.

Note: The pressure is considered to be one atm, not one bar if one mole of a gas occupies $22.4$ L at NTP. Standard conditions of temperature and pressure are often used to define the standard reference conditions to express volumes of liquids and gases. Also, nitrous oxide is called laughing gas.