JAMB - Chemistry (2004 - No. 7)
A given volume of methane diffuses in 20s. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions?
[C = 12, H = 1, S = 32, O = 16
[C = 12, H = 1, S = 32, O = 16
5s
20s
40s
60s
Explanation
Applying the Graham's law definition; i.e,
\(\frac{RSO_2}{RCH_4}\) = \(\sqrt{\frac{MCH_4}{MSO_2}}\)
\(\frac{tSO_2}{tCH_4}\) = \(\sqrt{\frac{MSO_2}{MCH_4}}\) (rate of diffusion ∝ \(\frac{1}{time taken}\))
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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