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JEE Advanced levelSingle correct optionDifficulty 3 of 3ig093

A gas has an rms speed of $412 m s^{- 1}$ at $300 K$ . Taking $R = 8.314 J K^{- 1} mo l^{- 1}$ , the gas is most likely

A $Ar$ (40)
B $N X_{2}$ (28)
C $CO X_{2}$ (44)
D $SO X_{2}$ (64)

Answer: C

M = \frac{3 RT}{u ^{2}} = \frac{3 \times 8.314 \times 300}{41 2 ^{2}} = 0.0441 kg mo l^{- 1} = 44.1 g mo l^{- 1}

, which matches

CO X_{2}

(44).

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