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Q.
The root mean square velocity of gas molecules at $0^\circ C$ will be if at N.T.P. its density is $1.43\,Kg /m^3$
Kinetic Theory
Solution:
According to formula
$\Rightarrow P =\frac{1}{3} \rho \bar{ C }^{2}$
Where $P$ is pressure of gas
$\bar{ C }$ is rms velocity
$\rho$ is density of gas
Given,
$\Rightarrow P =1 \,atm =1.013 \times 10^{5} $ pascal
$\Rightarrow \rho=1.43\, kg / m ^{3} $
So, $\bar{ C }=\sqrt{\frac{3 P }{\rho}}=\sqrt{\frac{3 \times 1.013 \times 10^{5}}{1.43}}$
$\Rightarrow \overline{ C } \cong 461 \,m / s$
Hence, the answer is $461 \,m / s$.