What observed rotation is expected when a 1.30 M solution of (R)-2-butanol is mi
ID: 511725 • Letter: W
Question
What observed rotation is expected when a 1.30 M solution of (R)-2-butanol is mixed with an equal volume of a 0.650 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.
What observed rotation is expected when a 1.30 M solution of (R)-2-butanol is mixed with an equal volume of a 0.650 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 dm 1 Number deg AC Previous Check Answer O Next Exit Hint The equation for observed optical rotation is [a] cl When performing the calculation, be sure that the units cancel to give degreesExplanation / Answer
The molecular mass of 2 butanol=74.122 g/mole.
The concentration of 1.30 M (R)-2-butanol = 74.122 x 1.3 g L-1 = 0.09635 g ml-1.
The optical activity of any racemic mixture = 0
Hence, the optical activity of 1.30 M solution of (R)-2-butanol remains unchanged upon mixing with an equal volume of a 0.650 M solution of racemic 2-butanol.
Therefore, the observed rotation of the solution is equal to the observed rotation of 1.30 M (R)-2-butanol.
Observed rotation (degrees) = Specific rotation (degrees mL g–1 dm–1) x concentration (g ml-1) x length (dm)
= +13.5 (degrees mL g–1 dm–1) x 0.0963586 (g ml-1) x 1(dm)
= 1.339 degrees.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.