AP EAMCET2010ChemistryStates of Matter
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)
Options
- AN
- BN 2
- CN 4
- DN 16
Correct answer
B. N 2
Step-by-step solution
From Graham's law of diffusion, r_ CH ₄ r_X = M_X M_ CH ₄ (given, r_ CH ₄ =2 r_X ) 2= M_X 16 M_X=16 4=64 Thus, the molecular mass of gas X is 64 . Number of molecules of gas X in 32 ~g gas = 32 64 N= N 2