170k views
1 vote
A 0.150 m solution of an enantiomerically pure chiral compound d has an observed rotation of 0.18 degrees in a 1-dm sample container. the molar mass of the compound if 176.0g/mol. what is the specific rotation of d?

User Beowulf
by
7.1k points

1 Answer

5 votes

Final answer:

The specific rotation of the chiral compound 'd' is calculated to be 6.82° · mL/g · dm using the observed rotation of 0.18 degrees, the path length of 1 dm, the molar mass of 176.0 g/mol, and the molarity of 0.150.

Step-by-step explanation:

The student is asking about how to calculate the specific rotation of a chiral compound known as d. The specific rotation is found using the formula α = [α]D(lc), where α is the observed rotation, [α]D is the specific rotation, l is the path length in decimeters, and c is the concentration of the solution in grams per milliliter. Given the observed rotation (α) is 0.18 degrees, the path length (l) is 1 dm, and the concentration is 0.150 molarity in a 1-dm sample container, we can calculate the specific rotation [α]D. First, we must convert the concentration to g/mL using the molar mass (176.0 g/mol).

To find specific rotation, the concentration needs to be in grams per milliliter. The molar concentration (0.150 m) is multiplied by the molar mass (176.0 g/mol) to get grams per liter, which is then converted to grams per milliliter by dividing by 1000. This gives us the concentration (c) in the desired units:

c = 0.150 mol/L × 176.0 g/mol ÷ 1000 mL/L = 0.0264 g/mL

Then we can calculate the specific rotation ([α]D) using the formula:

[α]D = α / (lc) = 0.18° / (1 × 0.0264) = 6.82° · mL/g · dm

Therefore, the specific rotation of compound d is 6.82° · mL/g · dm.

User Aage Torleif
by
7.9k points