Question

Geraniol, C10H17OH, is an rose-scented alcohol commonly used in perfumes. How many grams of geraniol should be added to 600. mL of ethanol in order to have a solution that boils at 82.00°C? [For ethanol, Kb = 1.22 °C/m, density = 0.789 g/cm3, boiling point = 78.40°C]

Answer 1

Answer:

215.12 g

Explanation:

The change in the bolling temperature is a collateral property called ebullioscopy. The elevation on temperature occurs because there'll be a solute in the pure compound, and that solute(generally non-volatile) has a different boiling point, so the total bolling pint should increase.

To calculate the elevation in the temperature, we must use the Raoult equation:

ΔT = KbxWxi

Where ΔT is the variation of the temperature of boiling point, Kb is the

ebullioscopy constant, W is the molarity, and i the Vant' Hoff factor, which for molecules is also equal to 1, because they don't form ions. So:

82.00 - 78.40 = 1.22xW

W = 3.60/1.22

W = 2.9508

The molarity can also be calculated by:

$W = \frac{m1}{M1xm2}$

Where m1 is the mass of the solute (in grames), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).

The density of ethanol is 0.789 g/cm³, and 1 mL = 1 cm³, so 600.00 mL = 600.00 cm³ :

0.789 = m2/600

m2 = 473.4 g = 0.4734 kg

The molar mass of geraniol is:

C: 12 g/mol x 10 = 120 g/mol

H: 1 g/mol x 18 = 18 g/mol

O: 16 g/mol x 1 = 16 g/mol

M1 = 154 g/mol

Then:

$2.9508 = \frac{m1}{154x0.4734}$

m1 = 215.12 g of geraniol.