(R)-X has $[α] = - 48.0$ and (S)-X has $[α] = + 48.0$ (units degree dm $^{- 1}$ (g/mL) $^{- 1}$ ). A solution contains 0.12 g/mL of (R)-X and 0.04 g/mL of (S)-X and is examined in a 2.0 dm tube. The observed rotation is

Answer: C

The rotations add:

α = (- 48.0) (2.0) (0.12) + (+ 48.0) (2.0) (0.04) = - 11.52 + 3.84 = - 7.6 8^{\circ}

. Ignoring the (S) component gives

- 11.5 2^{\circ}

, and omitting the tube length gives

- 3.8 4^{\circ}