Identify the correct IUPAC name of  hydrocarbon (x) containing three  primary carbons atoms and with molar mass $$72$$ g mol$$^{-1}$$ :

Step 1:

For a saturated hydrocarbon (alkane), the general formula is $$C_nH_{2n+2}$$.

Molar mass is $$72g/mol^{-1}$$

Therefore,

$$12n+(2n+2)=72$$
$$14n+2=72$$
$$14n=70$$
$$n=5$$

the molecular formula is $$C_5H_{12}$$

Step 2:

Now check its isomers:

1. $$n-pentane$$ → 2 primary carbons
2. $$2-methylbutane$$ → 3 primary carbons (given in question)
3. $$2,2-dimethylpropane$$ → 4 primary carbons

Therefore, the IUPAC name of the compound with 3 Primary Carbons and $$72 g/mol^{-1}$$ mass is 

$$2-methylbutane$$ 

$$i.e$$ Option C.