If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]
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Correct Answer: Option B
Explanation:
Rate of diffusion \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Mm}}\)
Rate = \(\frac{1}{time}\)
\(\frac{1}{time}\) \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Molarmass}}\)
Time \(\alpha \sqrt{Density}\) or \(\sqrt{Molarmass}\)
At constant volume of 100cm3
\(\frac{t_{o2}}{\sqrt{Mn_{o2}}}\) = \(\frac{t_{n2}}{\sqrt{1 \times 2}}\)
\(\frac{50}{4 \sqrt{2}}\) = \(\frac{t_{n2}}{\sqrt{2}}\)
tn2= 12.5s
Rate of diffusion \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Mm}}\)
Rate = \(\frac{1}{time}\)
\(\frac{1}{time}\) \(\frac{\alpha_1}{Density}\) or \(\frac{1}{\sqrt{Molarmass}}\)
Time \(\alpha \sqrt{Density}\) or \(\sqrt{Molarmass}\)
At constant volume of 100cm3
\(\frac{t_{o2}}{\sqrt{Mn_{o2}}}\) = \(\frac{t_{n2}}{\sqrt{1 \times 2}}\)
\(\frac{50}{4 \sqrt{2}}\) = \(\frac{t_{n2}}{\sqrt{2}}\)
tn2= 12.5s