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$CH_4$ diffuses two times faster than a gas X. The number of molecules present in 32 g of gas X is : (N is Avogadro number)

\[\begin {array} {1 1} (1)\;N & \quad (2)\;\frac{N}{2} \\ (3)\;\frac{N}{4} & \quad (4)\;\frac{N}{16} \end {array}\]

1 Answer

$(2)\;\frac{N}{2}$
answered Nov 7, 2013 by pady_1
 

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