Question

1. calculate the specific rotation of (2r,3r)-tartaric acid based on the following observation: a 0.856 - g sample of the pure acid was diluted to 10.0 ml with water and placed in a 1.00 dm polarimeter tube. the observed rotation using the 589 nm line of a sodium lamp at 20.0°c was 1.06°.

Explanation:

Step1: Calculate the concentration \( c \)

Concentration \( c=\frac{\text{mass of sample (g)}}{\text{volume of solution (mL)}}\)
\(c = \frac{0.856\space g}{10.0\space mL}=0.0856\space g/mL\)

Step2: Use the formula for specific rotation \([\alpha]_D^T\)

The formula for specific rotation is \([\alpha]_D^T=\frac{\alpha}{l\times c}\)
where \(\alpha\) is the observed rotation (\(\alpha = 1.06^{\circ}\)), \(l\) is the path - length in decimeters (\(l = 1.00\space dm\)), and \(c\) is the concentration in \(g/mL\)

Substitute the values into the formula:
\([\alpha]_D^{20}=\frac{1.06^{\circ}}{1.00\space dm\times0.0856\space g/mL}\)

Step3: Calculate the specific rotation

\([\alpha]_D^{20}=\frac{1.06}{0.0856}\space ^{\circ}mL/(g\cdot dm)\approx12.4^{\circ}mL/(g\cdot dm)\)

Answer:

The specific rotation of \((2R, 3R)\) - tartaric acid is approximately \(12.4^{\circ}\) (with units \(^{\circ}mL/(g\cdot dm)\))