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

Weight of 2L2{\text{L}} of Nitrogen at NTP is:
A.2.5g2.5{\text{g}}
B.1.25g1.25{\text{g}}
C.2.33g2.33{\text{g}}
D.14.0g14.0{\text{g}}

Explanation

Solution

One mole of any gas at NTP occupies 22.4L22.4{\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 =293.15K temperature and 1atm pressure = 293.15{\text{K temperature and }}1{\text{atm pressure}} and STP stands for standard temperature and pressure =273K temperature and 1bar 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 12g12{\text{g}} of carbon (C12)\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 NA{{\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 22.4L22.4{\text{L}} of any gas occupied at NTP condition.
It can be written as:
no of mole=given mass(in gram)molar mass(in gmol−1)=given volume at NTP(in litres)22.4L=given number of particles6.022×1023{\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 2L2{\text{L}}, so number of moles of Nitrogen present at NTP will be:

no of mole=given volume at NTP(in litres)22.4L=2L22.4L=0.089mol{\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 28gmol−128{\text{gmo}}{{\text{l}}^{ - 1}}.
Mass of nitrogen will be: mass of Nitrogen=number of moles×molar mass of Nitrogen{\text{mass of Nitrogen}} = {\text{number of moles}} \times {\text{molar mass of Nitrogen}}
⇒Mass of Nitrogen=0.089×28=2.5g\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.