Question

A 0.110 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.24° in a 1-dm sample container. The molar mass of the compound is 144.0 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 0.110 M in L, the enantiomer of D? Number deg mL Number g dm deg (c) What is the observed rotation if the solution of D is diluted with an equal volume of solvent? (d) What is the specific rotation of D after the dilution described in part (c)? Number Number deg mL deg p . (f) What is the observed rotation of 100 mL of a solution that contains 0.01 mole of D and 0.005 mole of L? (Assume a 1-dm path length.) (e) What is the specific rotation of L, the enantiomer of D, after the dilution described in part (c)? Number Number deg mL deg gdm

Answer 1

Solution : Given that : ca concentration = 0.10 X 144.0 2 15.84 g/ml. ob Soved relation -+0.one Specific notation = obseved rotation Ix concentration 0-24 *15084 FO24'X 15.84 Specific rotation. .= (3.2016) degree.ml/g.dm . observed rotation to degrey c=15. on & 7292 choved rotation = 3.801677.98 xl observed rotation - 30.10) degree

d Specific vctarion 18 ineffected by dilesion Specific Yotation = 3.8016 © Specific rotation of L = -3.8016. moles = 0.01 -0.005 . = 0.005 cum Concentration = 0.005 Oil = 0.05M -0.05 X 144 Lond = 7.2 200 - lo.corp. 0.0072 x 3.8016 Obs Ned Yotation observed rotation = 0.027

Answer 2

a.) person above forgot to convert 15.84g/L to g/mL which is 0.01584g/mL so when using that number the specific rotation is 0.24/0.01584g/mL = 15.1

b.) zero

c.) use the correct numbers to plug in from a.) so 0.01584g/mL / 2 = 0.00792M

0.00792M * 15.1 = ???

d.) dilition doesn't affect specific rotaion so = 15.1

e.) enantiomer so oppsoite sign = -15.1

f.) do it like above except plug in the correct roation of 15.1 instead 3.8016