Question: A 0.100 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 compount is 142.0 g/mol.
(a) What is the specific rotation of D?
Attempting to work out the problem: So I understand that I'm using the formula [specific rotation = observed rotation / (c*l)]. So here is how I tried to find c:
.100 M = mol/L
142 g/mol
.1*142 =14.2 g/L* 1 L/ 1000 mL = .0142
So the equation would me +0.24/(.0142)(1)...
...making the specific rotation 16.9014
(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?
Attempt: Does this mean I add an additional .100 M to my equation?
Sort of like:
16.9014 = observed rotation/ 2(.0142)
I don't quite understand the question and would appreciate if you could help me out.
I thank you in advance.