Consider the isomers of hydrocarbon with molecular formula C_5H_ 10...

The hydrocarbon with molecular formula $C_5H_{10}$ has a degree of unsaturation of $1$. Since its isomers do not decolourise $KMnO_4$ solution, they do not contain any carbon-carbon double bonds. Thus, they must be cycloalkanes.

The possible structural isomers of cyclic $C_5H_{10}$ and the number of their monochloro structural isomers are as follows:

1. Cyclopentane:
All carbon atoms are equivalent. Chlorination gives $1$ monochloro product (chlorocyclopentane).

2. Methylcyclobutane:
There are $4$ different types of hydrogen atoms (on the methyl group, on $C_1$, on $C_2/C_4$, and on $C_3$). Chlorination gives $4$ monochloro products.

3. Ethylcyclopropane:
There are $4$ different types of hydrogen atoms (on the $CH_3$ of ethyl, on the $CH_2$ of ethyl, on $C_1$ of the ring, and on $C_2/C_3$ of the ring). Chlorination gives $4$ monochloro products.

4. 1,1-Dimethylcyclopropane:
There are $2$ different types of hydrogen atoms (on the two equivalent methyl groups, and on the two equivalent $CH_2$ groups of the ring). Chlorination gives $2$ monochloro products.

5. 1,2-Dimethylcyclopropane:
There are $3$ different types of hydrogen atoms (on the two equivalent methyl groups, on the two equivalent $CH$ groups of the ring, and on the $CH_2$ group of the ring). Chlorination gives $3$ monochloro products.

Since each of these cycloalkanes has a different carbon skeleton, their monochloro derivatives will all be distinct structural isomers.

Total number of monochloro compounds (structural isomers only) = $1 + 4 + 4 + 2 + 3 = 14$.

Answer: $14$