At what temperature will average speed of the molecules of the second member of the series `C_(n)H_(2n)` be the same to that of `Cl_(2)` at `627^(@)C`?
A
259.4 K
B
400 K
C
532.4 K
D
None of these
Text Solution
Verified by Experts
The correct Answer is:
c
(c) Second member of `C_(n)H_(2n)` series `=C_(3)H_(6)=42` `=sqrt((8RT_(1))/(piM_(1)))=sqrt((8RT_(2))/(piM_(2)))=(900)/(71)=(T_(2))/(42)` `T_(2)=532.4K`
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