An alkane (Y) requires 8 moles of oxygen for complete combustion...

The general formula of an alkane is $C_n H_{2n+2}$.

The balanced chemical equation for the complete combustion of an alkane is:

$C_n H_{2n+2} + \left(\dfrac{3n+1}{2}\right) O_2 \rightarrow n CO_2 + (n+1) H_2O$

Given that $8$ moles of oxygen are required for complete combustion:

$\dfrac{3n+1}{2} = 8$

$3n + 1 = 16$

$n = 5$

The molecular formula of the alkane $(Y)$ is $C_5H_{12}$.

Since $(Y)$ yields only one monochlorinated product $(Z)$ upon reaction with $Cl_2/h\nu$, all the hydrogen atoms in the alkane must be equivalent.

Among the isomers of $C_5H_{12}$ (n-pentane, isopentane, and neopentane), only 2,2-dimethylpropane (neopentane) has all equivalent hydrogen atoms.

The structure of neopentane is $C(CH_3)_4$.

In this structure, the central carbon is quaternary, and the four surrounding methyl carbons are primary carbon atoms.

Therefore, the total number of primary carbon atoms in $(Y)$ is $4$.

Answer: $4$