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{:(,"Column-I",,"Column-II"),("A)",N(2),...

`{:(,"Column-I",,"Column-II"),("A)",N_(2),"P)","40% carbon by mass"),("B)",CO,"Q)","Empirical formula "CH_(2)O),("C)",C_(6)H_(12)O_(6),"R)","Vapour density : 14"),("D)",CH_(3)COOH,"S)",14N_(A)" electrons in a mole"):}`

Text Solution

Verified by Experts

The correct Answer is:
A-RS; B-RS; C-PQ; D-PQ
Vapour density of nitrogen = 14
1 Mole nitrogen `(N_(2))=6.23xx10^(23)` molecules
`=14xx6.023xx10^(23)` electrons
Vapour density of CO = 14
1 mole carbon monoxide (CO)
`=6.023xx10^(23)` molecules
`=14xx6.023xx10^(23)` electrons
`C_(6)H_(12)O_(6)%` carbon by mass = `(72)/(180)xx100=40`
Empirical formula = `CH_(2)O`
`CH_(3)COOH%` carbon by mass
`(24)/(60)xx100=40`
Empirical formula = `CH_(2)O`
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