Sulphur (IV) oxide travels a given distance in \(10 sec\). How long will it take equal volume of helium to travel the same distance under the same conditions? \([ S =32, O =16, He\) \(=4\) ]
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Correct Answer: Option D
Explanation:
\(T_{ SO _{2}}=10 sec , T_{l l e}=\) ?
\(MMSO _{2}=64 g / mol\)
\(R \alpha \frac{1}{\sqrt{M M}} \Rightarrow \frac{R SO _{2}}{R_{\text {He }}}=\sqrt{\frac{M M H e}{M M S O_{2}}}\)
Rate \(=\frac{\text { distance }}{\text { time }}\) or \(\frac{\text { vol }}{\text { time }}\);
\(\frac{d_{ SO _{3}}}{T_{ SO _{2}}} \div \frac{d_{H e}}{T_{H e}}=\sqrt{\frac{4}{64}}\)
\(=\frac{d_{ SO _{1}}}{T_{ SO _{2}}} \times \frac{T_{ He }}{d_{ He }}=\sqrt{\frac{1}{16}} ;\)
\(d_{ SO _{2}}=d_{ He }\) ?
\(\frac{T_{\text {He }}}{10}=\frac{1}{4}, T_{H e}=\frac{10}{4}=2.5 sec\)
\(T_{ SO _{2}}=10 sec , T_{l l e}=\) ?
\(MMSO _{2}=64 g / mol\)
\(R \alpha \frac{1}{\sqrt{M M}} \Rightarrow \frac{R SO _{2}}{R_{\text {He }}}=\sqrt{\frac{M M H e}{M M S O_{2}}}\)
Rate \(=\frac{\text { distance }}{\text { time }}\) or \(\frac{\text { vol }}{\text { time }}\);
\(\frac{d_{ SO _{3}}}{T_{ SO _{2}}} \div \frac{d_{H e}}{T_{H e}}=\sqrt{\frac{4}{64}}\)
\(=\frac{d_{ SO _{1}}}{T_{ SO _{2}}} \times \frac{T_{ He }}{d_{ He }}=\sqrt{\frac{1}{16}} ;\)
\(d_{ SO _{2}}=d_{ He }\) ?
\(\frac{T_{\text {He }}}{10}=\frac{1}{4}, T_{H e}=\frac{10}{4}=2.5 sec\)