Match the hydrocarbons in Column I with the boiling points given in Column II.
Column I |
Column II |
(i) n–Pentane |
(a) 282.5 K |
(ii) iso-Pentane |
(b) 309 K |
(iii) neo-Pentane |
(c) 301 K |
Explanation:
(i) There are more Vander Waal’s forces in n-pentane, and its boiling point is also high since there is no branching and surface area.
(ii) The boiling point of iso-pentane is less because the molar mass is the same except there’s one brach which reduces the surface area.
(iii) The boiling point of neo-pentane is the lowest because it has two side chains which have the same molar mass.