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The mean speed of Helium atoms is twice that of the molecules of a gas ‘X’ at the same temperature. The molecular weight of gas ‘X’

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answer is B.

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Detailed Solution

$\large {\left( {{V_{avg}}} \right)_{He}} = 2{\left( {{V_{avg}}} \right)_X}$

$\large \sqrt {\frac{{8RT}}{{\pi M}}} = 2\sqrt {\frac{{8RT}}{{{\pi _X}{M_X}}}}$

$\large \frac{1}{4} = \frac{4}{{{M_X}}} \Rightarrow {M_X} = 16\$

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