A sample of Buna-S contains buta-1,3-diene ( $C X_{4} H X_{6}$ , 54) and styrene ( $C X_{8} H X_{8}$ , 104) residues in the mole ratio $3 : 1$ . Taking the mass of each residue equal to the monomer mass, the mass percentage of styrene in the copolymer is

Answer: D

Per 4 residues: butadiene

= 3 \times 54 = 162

, styrene

= 104

, total

266

. Styrene mass percent

= 104/266 \times 100 = 39.1%

. The value

25.0%

is the mole percent, and

60.9%

is the butadiene share.