A 250 mL solution contains a partly resolved sample of compound Y at 0.0500 g/mL. In a 2.00 dm tube the observed rotation is -3.60 deg. Pure (-)-Y has [alpha] = -90.0 (degree dm^-1 (g/mL)^-1). The mass of the (+) enantiomer present in the solution is

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A 250 mL solution contains a partly resolved sample of compound Y at 0.0500 g/mL. In a 2.00 dm tube the observed rotation is $- 3.6 0^{\circ}$ . Pure (-)-Y has $[α] = - 90.0$ (degree dm $^{- 1}$ (g/mL) $^{- 1}$ ). The mass of the (+) enantiomer present in the solution is

A3.75 g
B8.75 g
C5.00 g
D7.50 g

Answer: A

[α]_{o b s} = α / (l c) = - 3.60/ (2.00 \times 0.0500) = - 36.0

. ee = 36.0/90.0 = 40%, so (-) = 70% and (+) = 30%. Total mass = 0.0500 x 250 = 12.5 g, so (+) = 0.30 x 12.5 = 3.75 g. The value 8.75 g is the (-) enantiomer.

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A 250 mL solution contains a partly resolved sample of compound Y at 0.0500 g/mL. In a 2.00 dm tube the observed rotation is −3.60∘. Pure (-)-Y has [α]=−90.0 (degree dm−1 (g/mL)−1). The mass of the (+) enantiomer present in the solution is

Answer: A

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A 250 mL solution contains a partly resolved sample of compound Y at 0.0500 g/mL. In a 2.00 dm tube the observed rotation is $- 3.6 0^{\circ}$ . Pure (-)-Y has $[α] = - 90.0$ (degree dm $^{- 1}$ (g/mL) $^{- 1}$ ). The mass of the (+) enantiomer present in the solution is