A 0.175 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.27° in a 1-dm sample container. The molar mass of the compound is 133.0g/mol.
What is the specific rotation of D?

The specific rotation of D is 11.60° mL/g dm

Explanation:

Given that:

The path length (l) =  1 dm

Observed rotation (∝) = + 0.27°

Molarity = 0.175 M

Molar mass = 133.0 g/mol

Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol

Concentration in (g/mL) = 23.275 g/L

Since 1 L = 1000 mL

Concentration in (g/mL) = 0.023275 g/mL

The specific rotation [∝] = ∝/(1×c)

= 0.27°/( 1  dm ×  0.023275 g/mL )

= 11.60° mL/g dm

Thus, the specific rotation of D is 11.60° mL/g dm

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )