## Answers

Mass of retinol = 153.0 mg = 0.153 g

Mass of solvent ( Chloroform) = 1.000 g = 0.001 kg

We have relation, B.P of solution - B.P of solvent = K _{b} x m

Where , K _{b} is boiling point constant of solvent , m is molality of solution

1.939 ^{0} C = 3.630 ^{0} C /m x m

m = 1.939 ^{0} C / 3.630 ^{0} C /m = 0.5342 m

We know that, molality = No. of moles of solute/ Mass of solvent in kg = (Mass / Molar mass ) of solute / Mass of solvent in kg

Therefore, Molality of retinol solution = (Mass / Molar mass ) of retinol / Mass of chloroform in kg

0.5342 m = ( 0.153 g / Molar Mass ) / 0.001 kg

( 0.153 g / Molar Mass ) = 0.5342 mol / kg x 0.001 kg = 5.342 x 10 ^{-04} mol

Molar Mass of retinol = 0.153 g / 5.342 x 10 ^{-04} mol = 286.41 g / mol

**ANSWER : Molar mass of retinol = 286.41 g / mol**

To determine molecular formula of retinol we must know the moles of constituent elements present in retinol.

Moles of C = % of C/ Atomic mass of C = 83.84 / 12.01 = 6.981

Moles of H = % of H / Atomic mass of H = 10.58 / 1.0079 = 10.50

Moles of O = % of O / Atomic mass of O = 5.58 / 16.00 = 0.3488

Hence, the ratio of number of moles of C:H:O is

(6.981 / 0.3488 = 20 , 10.50 / 0.3488 = 30 and 0.3488 / 0.3488 = 1)

Hence, empirical formula is C_{20}H_{30}O

Empirical formula mass = ( 20 x 12.01) + ( 30 x 1.0079) + 16.00 = 286.24

Therefore, ( molar mass / empirical formula mass ) = 286.41 / 286.24 = 1

i e molar mass = empirical mass

Therefore, Molecular formula = empirical formula = C_{20}H_{30}O

**ANSWER : Molecular formula of retinol = C _{20}H_{30}O**