Question

iii. Deduce the mass of Eugenol in 1000 liters of distillate. Eugenol, Mr = 164 Da. moi Specific Gravity = 1.06

from previous parts, the boiling point of the oil/steam is 84C and molar composition of Eugenol is 0.4534 at this point

Answer 1

Let total moles given are 1 mol out of which moles of Eugenol will be 0.4534 and moles of water = 1–0.4534 = 0.5466

weight of water = mol x molar mass = 0.5466 mol x 18 gm mol^—1 = 9.84 gm

weight of Eugenol = 0.4534 mol x 164 gm / mol =74.36 gm

total mass = 9.84 + 74.36 = 84.2 gm

volume of water = mass / density

density = 1.06 gm / mL

volume = 84.2 gm / 1.06 gm mL^—1 = 79.43 mL

molarity = moles of solute / volume of solution in litre = 0.4534 mol x (1000 / 79.4 litre ) = 5.71 M