hellppp please i have a test on this on mondayy im so screwed

Question

A solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25°C is observed to be 0.4010 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution.)

Answer 1

Osmotic pressure is given by:
π = MRT
π = osmotic pressure in atm
Mi = moles of particles/liter <- unknown
R = 0.0821 L.atm/K.mol
T = 298 K

• Substitute and solve for Mi to moles of ions per liter.
• Divide M by 3 to get the overall molarity, M, in terms of MgCl2 formula units.
• Multiply M by the molar mass of MgCl2 to get grams / liter which is equal to grams/(1000mLs)
• Divide by 10 to get grams / 100 mLs = ___%

(NOTE: We assume the density of the solution is close to 1.00 g/mL, so that grams/100mLs is about the same as the mass % )

Answer 2

Osmotic pressure is given by:
π = MRT
π = osmotic pressure in atm
Mi = moles of particles/liter <- unknown
R = 0.0821 L.atm/K.mol
T = 298 K

• Substitute and solve for Mi to moles of ions per liter.
• Divide Mi by 3 to get the overall molarity, M, in terms of MgCl2 formula units/L.
• Multiply M by the molar mass of MgCl2 to get grams / liter which is equal to grams/(1000mLs)
• Divide by 10 to get grams / 100 mLs = ___%

(NOTE: We assume the density of the solution is close to 1.00 g/mL, so that grams/100mLs is about the same as the mass % )