Molar mass of the hydrocarbon (X) which on ozonolysis consumes one mole of O$$_3$$ per mole of (X) and gives one mole each of ethanal and propanone is ______ g mol$$^{-1}$$ (Molar mass of C: 12 g mol$$^{-1}$$, H: 1 g mol$$^{-1}$$)

Hydrocarbon (X) on ozonolysis consumes one mole of O₃ per mole of X and gives one mole each of ethanal (CH₃CHO) and propanone (CH₃COCH₃).

Since each C=C double bond consumes one mole of O₃ in ozonolysis, the consumption of one mole of O₃ indicates the presence of a single C=C double bond in the molecule.

Ozonolysis cleaves this double bond into two carbonyl fragments:

$$>C=C< \xrightarrow{O_3} >C=O + O=C<$$

One fragment is ethanal (CH₃CHO), implying one side of the original double bond was CH₃-CH=, and the other fragment is propanone ((CH₃)₂CO), implying the other side was =C(CH₃)₂.

Rejoining these fragments across the double bond produces the structure

$$CH_3\text{-}CH=C(CH_3)_2$$

which corresponds to 2-methylbut-2-ene.

The molecular formula of this hydrocarbon is C₅H₁₀, and its molar mass is

$$M = 5 \times 12 + 10 \times 1 = 60 + 10 = 70 \text{ g mol}^{-1}$$

Therefore, the molar mass of hydrocarbon X is 70 g mol⁻¹.