AP EAMCET · Chemistry · States of Matter
\(\mathrm{CH}_4\) diffuses two times faster than a gas \(X\). The number of molecules present in \(32 \mathrm{~g}\) of gas \(X\) is ( \(N\) is Avogadro number)
- A \(N\)
- B \(\frac{N}{2}\)
- C \(\frac{N}{4}\)
- D \(\frac{N}{16}\)
Answer & Solution
Correct Answer
(B) \(\frac{N}{2}\)
Step-by-step Solution
Detailed explanation
From Graham's law of diffusion, \(\frac{r_{\mathrm{CH}_4}}{r_X}=\sqrt{\frac{M_X}{M_{\mathrm{CH}_4}}}\) (given, \(r_{\mathrm{CH}_4}=2 \cdot r_X\) ) \(2=\sqrt{\frac{M_X}{16}}\) \(M_X=16 \times 4=64\) Thus, the molecular mass of gas \(X\) is 64 . Number of molecules of gas \(X\) in…
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