A hydrocarbon ‘X’ is found to have molar mass of $80$. A $10.0 mg$ of compound ‘X’ on hydrogenation consumed $8.40 mL$ of $H_2$ gas (measured at STP). Ozonolysis compound ‘X’ yields only formaldehyde and dialdehyde. The total number of fragments/molecules produced from the ozonolysis of compound ‘X’ is _

moles of  $X=\frac{10mg}{80}=0.125\text{\hspace{0.17em}m\hspace{0.17em}}mol.$
  $\begin{array}{l}Molesconsumedof{H}_2=\frac{8.4}{22.4}=0.375mmol.\\ \frac{{ }^n{H}_2}{n_X}=\frac{0.375}{0.125}=3\end{array}$
So, the compound X have 3 double bonds. Ozonolysis of the compound yields only formaldehyde and dialdehyde, so the compound should be

$\begin{array}{l}{H}_{2}C=CH-CH=CH-\stackrel{\stackrel{H}{|}}{C}=C{H}_2\to \underset{4molecules}{\underbracket{2HCHO+2C_2{O}_2{H}_2}}\\ Molecularmass=(12\times 6)+1\times 8=72+8=80amu\\ Ozonolysisform:\\ \end{array}$