Question:
What is the difference in molar masses of third and fourth homologues of alkane series?
What is the difference in molar masses of third and fourth homologues of alkane series?
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Successive homologues always differ by 14 mass units (one carbon + two hydrogens).
Updated On: Apr 30, 2026
- 28 g mol$^{-1}$
- 14 g mol$^{-1}$
- 15 g mol$^{-1}$
- 16 g mol$^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
In a homologous series, any two successive members differ by a $-CH_2-$ group.
Step 2: Analysis
- 3rd alkane: Propane ($C_3H_8$) - 4th alkane: Butane ($C_4H_{10}$)
Step 3: Calculation
Difference = Molar mass of $CH_2 = 12 (C) + 2 \times 1 (H) = 14$ g/mol.
Step 4: Conclusion
The difference is 14 g/mol.
Final Answer:(B)
In a homologous series, any two successive members differ by a $-CH_2-$ group.
Step 2: Analysis
- 3rd alkane: Propane ($C_3H_8$) - 4th alkane: Butane ($C_4H_{10}$)
Step 3: Calculation
Difference = Molar mass of $CH_2 = 12 (C) + 2 \times 1 (H) = 14$ g/mol.
Step 4: Conclusion
The difference is 14 g/mol.
Final Answer:(B)
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