Question

. 0.275 L of solution is prepared by dissolving 2.50 g of an unknown molecular compound in enough water. The osmotic pressure of this solution is 1.98 atm at 27 °C. (R = 0.08206 L·atm/mol·K) a) Determine the molar mass of the molecular compound. (7 pts) b) If you need to prepare 0.275 L of AlCl3 solution with the same osmotic pressure of the above solution, what mass of AlCl3 (MW = 133.33 g/mol) would you need to weigh out?

Answer 1

a) Molar mass = mRT/Vπ

Where T = 27 + 273.15 = 300.15 K

π = 1.98 atm

Molar mass, M = 2.50×0.08206×300.15/0.275×1.98

= 113.09 g/mol

b) Mass = πMV/RT

= 1.98×133.33×0.275/0.08206×300.15

= 2.95 g