Question

Weight of \[2{\text{L}}\] of Nitrogen at NTP is:
A.\[2.5{\text{g}}\]
B.\[1.25{\text{g}}\]
C.\[2.33{\text{g}}\]
D.\[14.0{\text{g}}\]

Answer 1

Hint: One mole of any gas at NTP occupies \[22.4{\text{L}}\] of volume. Mass of a substance is calculated by the product of the number of moles of that substance and its molar mass.

Complete step by step solution:
- NTP stands for normal temperature and pressure \[ = 293.15{\text{K temperature and }}1{\text{atm pressure}}\] and STP stands for standard temperature and pressure \[ = 273{\text{K temperature and }}1{\text{bar pressure}}\].
- Mole is the SI unit of amount of substance. It is defined as the amount of a substance that contains as many entities as there are atoms present in exactly \[12{\text{g}}\] of carbon \[\left( {{{\text{C}}^{12}}} \right)\].
- Mole is a concept of quantity in terms of number, mass and volume. For a given balanced equation, information of reactant or product can be determined if information of one of the species is given either in terms of moles, molecules or weight. 1 mole is equivalent to \[{{\text{N}}_{\text{A}}}\] atoms, molecules, ion, or electrons. 1 mole is equivalent to molecular weight or atomic weight of a substance. 1 mole is equivalent to the volume of \[22.4{\text{L}}\] of any gas occupied at NTP condition.
It can be written as:
\[{\text{no of mole}} = \dfrac{{{\text{given mass}}\left( {{\text{in gram}}} \right)}}{{{\text{molar mass}}\left( {{\text{in gmo}}{{\text{l}}^{ - 1}}} \right)}} = \dfrac{{{\text{given volume at NTP}}\left( {{\text{in litres}}} \right)}}{{22.4{\text{L}}}} = \dfrac{{{\text{given number of particles}}}}{{6.022 \times {{10}^{23}}}}\]
As given that volume of Nitrogen at is NTP \[2{\text{L}}\], so number of moles of Nitrogen present at NTP will be:

\[{\text{no of mole}} = \dfrac{{{\text{given volume at NTP}}\left( {{\text{in litres}}} \right)}}{{22.4{\text{L}}}} = \dfrac{{2{\text{L}}}}{{22.4{\text{L}}}} = 0.089{\text{mol}}\]
As we know, the molar mass of Nitrogen is \[28{\text{gmo}}{{\text{l}}^{ - 1}}\].
Mass of nitrogen will be: \[{\text{mass of Nitrogen}} = {\text{number of moles}} \times {\text{molar mass of Nitrogen}}\]
\[ \Rightarrow {\text{Mass of Nitrogen}} = 0.089 \times 28 = 2.5{\text{g}}\]

Thus, correct option is A.

Note: According to Avogadro law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.