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

 

 

Answers (1)

(i) \rightarrow (b); (ii) \rightarrow (c); (iii) \rightarrow (a)

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.

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