Isotopic variations in CHCl₃ produce distinct molecular masses

Steps:

• Assign masses: C = 12 u, H = 1 u, each Cl = 35 u or 37 u.
• Calculate base mass without Cl: 12 + 1 = 13 u.
• Determine Cl mass combinations for three Cl atoms: 3×35=105 u (all ^{35}Cl), 2×35+37=107 u, 35+2×37=109 u, 3×37=111 u.
• Add to base: 118 u, 120 u, 122 u, 124 u, yielding four distinct masses.

Why C is correct:

• Binomial expansion for three identical Cl atoms with two isotopes gives four unique mass values (n+1=4, where n=3).

Why the others are wrong:

• A: Ignores mixed isotope combinations, assuming only all-light or all-heavy.
• B: Misses one mixed combination, undercounting distinct masses.
• D: Overcounts by treating identical masses as distinct or adding nonexistent isotopes.

Final answer: C