At a certain temperature, the rms velocity for $ {O_2} $ is 400 m/s. at the same temperature, the rms velocity for $ {H_2} $ molecules will be
(A) 100 m/s
(B) 25 m/s
(C) 1600 m/s
(D) 6400 m/s

Hint: the root mean square velocity (rms velocity) is directly proportional to the square root of temperature and inversely proportional to the square root molecular mass of the gas. Since temperature is constant, find the constant as velocity times the square root of molar mass.

Formula used: In this solution we will be using the following formulae;
 $ {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where $ {v_{rms}} $ is the root mean square velocity (rms velocity), $ R $ is the universal gas constant, $ T $ is temperature and $ M $ is the molecular mass of the gas.

Complete Step-by-Step solution
To solve the above problem, we recall that the molar mass of oxygen is given as 32 g/mole. Now at a certain, unknown temperature, the rms velocity of oxygen is said to be 400 m/s.
Generally, the root mean square velocity of a gas can be given as
 $ {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where $ {v_{rms}} $ is the root mean square velocity (rms velocity), $ R $ is the universal gas constant, $ T $ is temperature and $ M $ is the molecular mass of the gas.
Hence, we can write for oxygen that
 $ 400 = \sqrt {\dfrac{{3RT}}{{32}}} $
This can be re-written as
 $ 400 = \dfrac{{\sqrt {3RT} }}{{\sqrt {32} }} $
Hence,
 $ \sqrt {3RT} = 400\sqrt {32} $
Now for hydrogen, we have that
 $ {u_{rms}} = \sqrt {\dfrac{{3RT}}{2}} $ (since the molecular mass of hydrogen is 2 g/mol) which can also be re-written as
 $ {u_{rms}} = \dfrac{{\sqrt {3RT} }}{{\sqrt 2 }} $
Hence, by inserting the constant, we have that
 $ {u_{rms}} = \dfrac{{400\sqrt {32} }}{{\sqrt 2 }} $
Which, through mathematical principles (the rule of surds), can be written as
 $ {u_{rms}} = \dfrac{{400\sqrt {2 \times } \sqrt {16} }}{{\sqrt 2 }} $
hence, by cancelling root of 2 and finding the square root of 16, then we have
 $ {u_{rms}} = 400 \times 4 = 1600m/s $
Hence, the correct option is C.

Note
For clarity, observe that we do not require to convert the unit from g/mol to kg/mol. This is because we did not insert any of the values of the constant as their SI unit, and the value of the temperature was not exactly found. Just the constant relation as velocity time molar mass. The molar mass, hence the unit will cancel out if we insert into the hydrogen rms velocity equation.