Question 21 of 25 rganic Map presented by Sapling Marc Loudon A 0.100 M solution

Question

Question 21 of 25 rganic Map presented by Sapling Marc Loudon A 0.100 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.12 in a 1-dm sample container. The molar mass of the compound is 125.0 g/mol. (a) What is the specific rotation of D? (b) What is the observed rotation if this solution is mixed with an equal volume of a solution that is 0.100 M in L, the enantiomer of D? Number deg mL Number g dma deg (c) What is the observed rotation if the solution (d) What is the specific rotation of D of Dis diluted with an equal volume of solvent? after the dilution described in part (c)? Number Number deg mL deg g dma (f) What is the observed rotation of 100 mL of a (e)What is the specific rotation of L, the enantiomer of solution that contains 0.01 mole of D and 0.005 D, after the dilution described in part (c)? mole of L? (Assume a 1-dm path length.) Number Number deg mL deg g dm AC Previous ® Give Up & View Solution Check Answer 0 Next Exit Hint

Explanation / Answer

(a)

Specific rotation = angle/(length in Dm)*(density in gm/ml) = 0.12/(1)*(1.25*10^ -2) =

9.6 deg.ml/g.dm

Density = Conc.*Molar mass = 0.1*125 = 12.5 g/L = 1.25*10^ -2 g/ml

(b) 0 deg

As the net rotation cancels due to the pressence of both D and L in same amount

(c)0.06 deg

As same amount of solvent is added Density becomes Half

Specific rotation is constant.

So the rotation also becomes Half

(d) +9.6 deg.ml/g.dm

Specific rotation do not change by change of conc. So it is same.

(e) -9.6 deg.ml/g.dm

Specific rotation of L is opposite of D at given temprature

(f) Observed Rotation = (Specific Rotation)*(length)*(density)

For D = (+9.6*(1)*(12.5*10^-3) = +0.12

For L = (-9.6)*(1)*(6.25*10^-3) = -0.06

Rotation = 0.12 -0.06 = +0.06 degree

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