What is the molar mass of third member of homologous series if the molar mass of first member is 46 g ?
What is the molar mass of third member of homologous series if the molar mass of first member is 46 g ?
60 g
74 g
138 g
80 g
Solution
Two successive homologues differ by one $-\mathrm{CH}_2$ - (methylene) unit, i.e. by a molar mass of $(12+2) \mathrm{g} \mathrm{mol}^{-1}=14 \mathrm{~g} \mathrm{~mol}^{-1}$.
Molar mass of first member $=46 \mathrm{~g}$
$\therefore \quad$ Molar mass of third member $=46+14+14$ $=74 \mathrm{~g}$