17 mg of a hydrocarbon (M.F. C$$_{10}$$H$$_{16}$$) takes up 8.40 mL of the H$$_2$$ gas measured at 0°C and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields

The number of double bond/s present in the hydrocarbon is

The amount of hydrogen consumed during hydrogenation can be used to determine the number of double bonds present in the hydrocarbon.

At STP, one mole of a gas occupies (22{,}400\ \text{mL}). Therefore, the number of moles of hydrogen used is

$$n_{H_2}=\frac{8.40}{22400}=3.75\times10^{-4}\ \text{mol}.$$

The molar mass of (C_{10}H_{16}) is

$$M=(10\times12)+(16\times1)=136\ \text{g mol}^{-1}.$$

Hence, the number of moles of hydrocarbon present is

$$n_{\text{hydrocarbon}}=\frac{0.017}{136}=1.25\times10^{-4}\ \text{mol}.$$

The number of moles of hydrogen consumed per mole of hydrocarbon is

$$\frac{n_{H_2}}{n_{\text{hydrocarbon}}}=\frac{3.75\times10^{-4}}{1.25\times10^{-4}}=3.$$

Since each mole of hydrogen saturates one double bond, the hydrocarbon contains

$$\boxed{3\ \text{double bonds}}.$$

This result is also consistent with the ozonolysis products obtained—acetone, two molecules of formaldehyde, and succinaldehyde—which collectively indicate the presence of three carbon-carbon double bonds in the original molecule.

Therefore, the hydrocarbon contains

$$\boxed{3\ \text{double bonds}}.$$