At room temperature $({27^0}C)$ ,the Root mean square speed of molecules of a certain diatomic gas is found to be $1920\,m/s$ . The gas is:
A. ${H_2}$
B. ${F_2}$
C. ${O_2}$
D. $C{l_2}$

Hint:Root mean square velocity of an atomic gas is defined as “the speed with which the molecules in the gas can travel at a particular temperature. At a given temperature, root mean square velocity is directly proportional to the absolute temperature and inversely proportional to the mass of that gas.

Complete answer:
The root mean square velocity of a gas gives us a brief idea about how fast the molecules of that gas are travelling at a particular temperature. As we have the formula for rms velocity,
${v_{rms}} = \sqrt {\dfrac{{3RT}}{m}} $
Where, $T$ is the temperature, $m$ Is the molar mass of the diatomic gas and $R$ Is the ideal gas constant which has a value of $8.314\,J{K^{ - 1}}{M^{ - 1}}$

In given question, we have $T = {27^0}C$ which we need to convert in Kelvin scale
$T = 27 + 273 \\
\Rightarrow T = 300K \\ $
${v_{rms}} = 1920\,m/s$
On putting these values in the formula ${v_{rms}} = \sqrt {\dfrac{{3RT}}{m}} $
We get, $1920 = \sqrt {\dfrac{{3 \times 8.314 \times 300}}{m}} $
$\sqrt m = \sqrt {3 \times 8.314 \times 300} \times \dfrac{1}{{1920}} \\
\therefore m = 0.002\,kgmo{l^{ - 1}}$
Hence, the diatomic gas molar mass found to be around $m = 2\,gmo{l^{ - 1}}$. Now, from given options the only gas is hydrogen whose molar mass is $2\,gmo{l^{ - 1}}$.

Hence, the correct option is A.

Note:We must remember about some conversions while solving these types of problem and the value of $R = $ $8.314J{K^{ - 1}}{M^{ - 1}}$ is a universal constant.
${0^0}C = 273K \\
\Rightarrow 1\,kg = 1000\,g$
And other gases like ${F_2}$ stands for Fluorine which has a molar mass of $37.99gmo{l^{ - 1}}$ , ${O_2}$ stands for Oxygen gas which has a molar mass of $32gmo{l^{ - 1}}$ , $C{l_2}$ stands for chlorine which has a molar mass of $70.90gmo{l^{ - 1}}$ .