WAEC - Chemistry (1997 - No. 10)
A given volume of methane diffuses in 20 seconds. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? [CH\(_4\) = 16; SO\(_2\) = 64]
5 seconds
20 seconds
40 seconds
60 seconds
80 seconds
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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