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`?

Options

A. 259.4K
B. 400 k
C. 532.4k
D. None of these

Answer

C

Explanation

Second member of `C_(n)H_(2n) ` series `= C_(3) H_(6) = 42`
`= sqrt(( 8 RT_(1))/( piM_(1)))= sqrt(( 8 RT_(2))/( piM_(2)))= ( 900)/( 71) = (T_(2))/( 42)`
`T_(2) = 532.4K`