Question

At NTP, sample of chlorine and oxygen is taken. Given that they both were of equal volumes, what will be the ratio of their number of molecules?
A. $1:1$
B. $32:27$
C. $2:1$
D. $16:14$

Answer 1

Hint:According to ideal gas law, $PV = nRT$ , where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles, $T$ is temperature and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . Use this law to calculate the required ratio.



Formula used:
Ideal gas law,
$PV = nRT$
Here $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles,
$T$ is the temperature and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .


Complete answer:
According to ideal gas law,
$PV = nRT$ … (1)
It is given that a sample of equal volume of chlorine and oxygen is taken.
Let this volume be $V$ .
It is also given that the sample is taken at NTP, the normal temperature and pressure.
Thus, both the temperature and pressure of the two gases are the same, that is, $293.15{\text{ K}}$ and $1{\text{ atm}}$ .
Let these values be $T$ and $P$ respectively.
Now, let the number of molecules of chlorine and oxygen be ${n_1}$ and ${n_2}$ respectively.
Then, using ideal gas law, given in formula (1),
Number of molecules of chlorine, ${n_1} = \dfrac{{PV}}{{RT}}$ … (2)
Similarly,
Number of molecules of oxygen, ${n_2} = \dfrac{{PV}}{{RT}}$ … (3)
Dividing (2) by (3), we get:
\[\dfrac{{{n_1}}}{{{n_2}}} = 1\]
Hence, the ratio of the number of moles and hence molecules of chlorine and oxygen is 1:1.
Thus, the correct option is A.



Note: According to Avogadro’s Postulate, all equal volumes of gases at the same temperature and pressure contain the same number of molecules or moles. You could have used this postulate to solve the above question as well.