The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of chlorophyll, C55H72MgN4O5, a nonvolatile, nonelectrolyte (MW = 893.5 g/mol), must be added to 187.4 grams of diethyl ether to reduce the vapor pressure to 457.45 mm Hg?

Question

Robert

Physics

1 October, 23:08

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of chlorophyll, C55H72MgN4O5, a nonvolatile, nonelectrolyte (MW = 893.5 g/mol), must be added to 187.4 grams of diethyl ether to reduce the vapor pressure to 457.45 mm Hg?

+3

Answers (1)

Know the Answer?

Harmony Watts · Answer 1

29.4855 grams of chlorophyll

Explanation:

From Raoult's law

Mole fraction of solvent = vapor pressure of solution : vapor pressure of solvent = 457.45 mmHg : 463.57 mmHg = 0.987

Mass of solvent (diethyl ether) = 187.4 g

MW of diethyl ether (C2H5OC2H5) = 74 g/mol

Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol

Let the moles of solute (chlorophyll) be y

Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.987 = 2.532 / (y + 2.532)

y + 2.532 = 2.532/0.987

y + 2.532 = 2.565

y = 2.565 - 2.532 = 0.033

Moles of solute (chlorophyll) = 0.033 mol

Mass of chlorophyll = moles of chlorophyll * MW = 0.033 * 893.5 = 29.4855 grams