CH4 diffuses two times faster than a gas X . The number of molecules present in 32 g of gas X is :( N is Avogadro number)
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Using graham's law of diffusion, rCH4rgas = √(mgasmCH4) 2 = √(mgas16) mgas = 64 Number of molecules = wmolar mass × N = 3264 × N = N2 .Hence, the ... More
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Answer:
CH4 diffuses two times faster than a gas X . The number of molecules present in 32 g of gas X is :( N is Avogadro number)
Top answer · 0 votes
Using graham's law of diffusion, rCH4rgas = √(mgasmCH4) 2 = √(mgas16) mgas = 64 Number of molecules = wmolar mass × N = 3264 × N = N2 .Hence, the ... More