Correct Answer: Option C

Explanation

Applying the Graham's law definition; i.e,

\(\frac{RSO_2}{RCH_4}\) = \(\sqrt{\frac{MCH_4}{MSO_2}}\) = \(\frac{tSO_2}{tCH_4}\)

Where t(SO\(_2\)) = ?, M(SO\(_2\)) = 64

t(CH\(_4\)) = 20 sec, M(CH\(_4\)) = 16

Therefore, \(\frac{t(SO_2)}{20}\) = \(\sqrt{\frac{64}{16}}\)

\(\frac{t(SO_2)}{20}\) = \(\frac{8}{4}\)

t(SO\(_2\)) = 8 x \(\frac{20}{4}\)

= 40 sec.

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