Sample A (6.0 g) has [alpha] = +30.0 and sample B (4.0 g) has [alpha] = -10.0, both samples of the same compound whose pure (+) enantiomer has [alpha] = +50.0 under the same conditions. The two samples are combined. The specific rotation of the 10.0 g mixture is

JEE Advanced levelSingle correct optionDifficulty 3 of 3st149

Sample A (6.0 g) has $[α] = + 30.0$ and sample B (4.0 g) has $[α] = - 10.0$ , both samples of the same compound whose pure (+) enantiomer has $[α] = + 50.0$ under the same conditions. The two samples are combined. The specific rotation of the 10.0 g mixture is

Excess (+) in A = 6.0 x (30.0/50.0) = 3.6 g; excess in B = 4.0 x (-10.0/50.0) = -0.8 g. Net excess = 2.8 g in 10.0 g, ee = 28.0%, so

[α] = 0.280 \times 50.0 = + 14.0

(equivalently the mass-weighted mean of the two specific rotations). The simple average is +10.0, which ignores the masses.