MHT CET · Chemistry · Haloalkanes and Haloarenes
What is the molar mass of product hydrocarbon when 2 moles of methyl bromide reacts with large excess of sodium in dry ether?
- A \(15.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(30.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(7.5 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(40.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(30.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation

Thus, the molar mass of the product formed, i.e., ethane is \(30.0 \mathrm{~g} \mathrm{~mol}^{-1}\).
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