If a pure R isomer has a specific rotation of –126.0°, and a sample contains 64.0% of the R isomer and 36.0% of its enantiomer, what is the observed specific rotation of the mixture?

The observed specific rotation of the mixture = 68.04°  

Explanation:

Given data

Specific rotation of the R isomer = - 126°

The bet rotation of the mixture is the rotation of excess enantiomer.

The R isomer = 64 %

Enantiomer = 36 %

The Enantiomer excess of the R isomer is given by

ee = 64 - 36 = 28 %

The observed specific rotation of the mixture is the rotation of this  Enantiomer excess.

Therefore the rotation of this  excess enantiomer

α = 0.54 × ( -126° )

α = 68.04°

Thus the observed specific rotation of the mixture = 68.04°  

its mg hope this

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"