
Number of molecules in $1$ litre of oxygen at NTP:
A. $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{32}}$
B. $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$
C. $32\, \times \,22.4$
D. $\dfrac{{32}}{{22.4}}$
Answer
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Hint: To answer this question we should know about the NTP. From the NTP condition we will compare the volume and number of moles to determine the number of moles of oxygen. Then by using Avogadro number we will compare the number of moles and number of molecules to determine the number of molecules of oxygen gas.
Complete step-by-step answer:
The NTP is known as normal temperature and pressure. The value of normal temperature in kelvin is$298\,{\text{K}}$. The value of the normal pressure in atm is${\text{1}}\,{\text{atm}}$. One mole of a gas at $298\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$volume.
So, from the NTP condition we know, $22.4$ L volume of oxygen is equal to one mole so, one liter of oxygen will be equal to,
$22.4$ L volume of a oxygen gas = one mole of the oxygen gas
$1$ L volume of a oxygen gas = $\dfrac{1}{22.4}$ mole of the oxygen gas
According to Avogadro number one mole of any substance contains $6.023\, \times \,{10^{23}}$ atoms ions or molecules.
From the Avogadro law, one mole of oxygen gas contains $6.023\, \times \,{10^{23}}$ oxygen molecules so, $1/22.4$mole of the oxygen gas will contains,
One mole of oxygen gas = $6.023\, \times \,{10^{23}}$Oxygen molecules
And we know,
One mole of oxygen gas = One oxygen molecule
So, $1/22.4$mole of the oxygen gas = $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$ oxygen molecules
So, the number of molecules in $1$ litre of oxygen at NTP is $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$.
Therefore, option (B) $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$ is correct.
Note: There is a slight difference between NTP and STP. STP is known as standard temperature and pressure. The value of standard temperature in kelvin is $273\,{\text{K}}$. The value of the standard pressure in atm is ${\text{1}}\,{\text{atm}}$. One mole of a gas at $273\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$ volume. NTP or STP condition gives the relation between the number of moles and volume.
Complete step-by-step answer:
The NTP is known as normal temperature and pressure. The value of normal temperature in kelvin is$298\,{\text{K}}$. The value of the normal pressure in atm is${\text{1}}\,{\text{atm}}$. One mole of a gas at $298\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$volume.
So, from the NTP condition we know, $22.4$ L volume of oxygen is equal to one mole so, one liter of oxygen will be equal to,
$22.4$ L volume of a oxygen gas = one mole of the oxygen gas
$1$ L volume of a oxygen gas = $\dfrac{1}{22.4}$ mole of the oxygen gas
According to Avogadro number one mole of any substance contains $6.023\, \times \,{10^{23}}$ atoms ions or molecules.
From the Avogadro law, one mole of oxygen gas contains $6.023\, \times \,{10^{23}}$ oxygen molecules so, $1/22.4$mole of the oxygen gas will contains,
One mole of oxygen gas = $6.023\, \times \,{10^{23}}$Oxygen molecules
And we know,
One mole of oxygen gas = One oxygen molecule
So, $1/22.4$mole of the oxygen gas = $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$ oxygen molecules
So, the number of molecules in $1$ litre of oxygen at NTP is $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$.
Therefore, option (B) $\dfrac{{6.023\, \times \,{{10}^{23}}}}{{22.4}}$ is correct.
Note: There is a slight difference between NTP and STP. STP is known as standard temperature and pressure. The value of standard temperature in kelvin is $273\,{\text{K}}$. The value of the standard pressure in atm is ${\text{1}}\,{\text{atm}}$. One mole of a gas at $273\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$ volume. NTP or STP condition gives the relation between the number of moles and volume.
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