The hydrocarbon ( X ) with molar mass 80 g mol − 1 and 90% carbon has …… degree of unsaturation.

∴ Empirical formula is C 3 H 4 Molecular mass = 80 g mol − 1 . Molecular formula = ( C 3 H 4 ) n n = EF mass Molecular mass = 40 80 = 2 ∴ Molecular formula is C 6 H 8 Degree of unsaturation = 2 2 × 6 + 2 − 8 = 3

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ChemistryConcentration TermsConcentration termEasy2 minPYQ_2025

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The hydrocarbon $(X)$ with molar mass $80 g mol^{- 1}$ and $90%$ carbon has $\dots\dots$ degree of unsaturation.

Answer:

Solution:

$∴$ Empirical formula is $C_{3} H_{4}$ Molecular mass $= 80 g mol^{- 1}$ . Molecular formula $= (C_{3} H_{4})_{n}$ $n = \frac{Molecular mass}{EF mass} = \frac{80}{40} = 2$ $∴$ Molecular formula is $C_{6} H_{8}$ Degree of unsaturation $= \frac{2 \times 6 + 2 - 8}{2} = 3$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Concentration term

⏱ 2mℹ️ Source: PYQ_2025

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