[Kayla_N]

1. Consider a solution that contains 64% R isomer and 36% S isomer. If the observed specific rotation of the mixture is -82.0 degrees, what is the specific rotation of the pure R isomer?

If you know that a 50:50 mixture = 0, and that a 64:36 mixture = -82, you can set up two equations to solve for the answer:
Let x = R isomer, and y = S isomer
0.5x + 0.5y = 0 (eq 1)
0.64x + 0.36y = -82 (eq 2)
solving for x and y in eq 1, we get y = -x
plugging this into equiation 2 we get:
0.64x + -0.36(-x) = -82
Combining the x terms we get:
0.28x = -82
and solving for x gives -292.85, the specific rotation for the R isomer
since y = -x, the rotation for the S isomer is 292.85

I was wondering if I do this right..if not please help out. thanks.

2. What observed rotation is expected when a 1.88 M solution of (R)-2-butanol is mixed with an equal volume of a 0.940 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).

Please help me with #2..i have no clue how to do it.