{:(,"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"):}

{:(,"Column-I",,"Column-II"),("A)",N_(2),"P)","40% carbon by mass"),("

1.	{:(,"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"):}
Answer» Solution :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`