Chemistry · Hydrocarbons

JEE Main 2025 — 24 January, Evening Shift — Question 45

The hydrocarbon (X) with molar mass 80 g mol−180 \mathrm{~g} \mathrm{~mol}^{-1} and 90%90 \% carbon has ___degree of unsaturation.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Mass of carbon =80×90100=72gm=\frac{80 \times 90}{100}=72 \mathrm{gm}

Number of C-atoms =7212=6=\frac{72}{12}=6

Mass of hydrogen =80×10800=8gm=\frac{80 \times 10}{800}=8 \mathrm{gm}

Number of H-atoms =81=8=\frac{8}{1}=8

So molecular formula C6H8\mathrm{C}_{6} \mathrm{H}_{8}

D.U. =6+1−8/2=7−4=3=6+1-8 / 2=7-4=3

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Hydrocarbons
Topic
Properties & Uses of Alkynes
The hydrocarbon (X) with molar mass 80 g mol -1 and 90 \% carbon has… | JEE Main 2025 PYQ with Solution · DhiX AI