Question

What observed rotation is expected when a 1.02 M solution of (R)-2-butanol is mixed with an equal volume of a 0.510 M solution of racemic 2-butanol, and the resulting solution is analyzed in a sample container that is 1 dm long? The specific rotation of (R)-2-butanol is –13.9 degrees mL g–1 dm–1.

Answer 1

1.02 M solution of (R)-2-butanol is mixed with an equal volume of a 0.510 M solution of racemic 2-butanol, in effect concentration of the (R)-2-butanol is halved. (no need to consider the racemic 2-butanol, since both the components are diluted and optical rotation will cancel each other)

The new concentration of the butanol is 0.51 M

molar mass of (R)-2-butanol is = 74.1 g/mol

since molarity is the number of moles of solute in 1000 mL solution.

Mass of (R)-2-butanol used to prepare 1000 mL = number of moles x molar mass = 74.1 x 0.51 = 145.29 g

concentration of the solution in g/mL = 145.29/1000 = 0.14529 g/mL

Specific rotation of the present sample = –13.9 degrees mL g–1 dm–1. x 0.14529 g/mLx 1 dm = 2.019 degrees