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? [CH4 = 16; SO2 = 64]
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Correct Answer: Option C
Explanation:
Applying the Graham's law defination; i.e,
\(\frac{R(SO_2)}{R(CH_4)} = \sqrt{\frac{M(CH_4)}{M(SO_2)}} = \frac{t(CH_4)}{t(SO_2)}\)
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.
Applying the Graham's law defination; i.e,
\(\frac{R(SO_2)}{R(CH_4)} = \sqrt{\frac{M(CH_4)}{M(SO_2)}} = \frac{t(CH_4)}{t(SO_2)}\)
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.