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(0.0034 / 0.035) times 100 Percent ionization = 9.7% A solid mixture contains both MgCl2 and NaCl. When 0.5000 g of this mixture is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25.0 Degree C is observed to be 0.3990 atm. What is the mass percent of MgCl2 in the solid? (Assume ideal behavior for the solution and that both ionic compounds dissociate completely in the water.)

Answer 1

0.3990 atm = Molarity x 0.08206 x 298.15

Molarity = 0.01631

MgCl2 yields 3 particles per molecule

NaCl yield 2 particles per molecule

MgCl2: Molar mass = 95.211 g

NaCl: Molar mass = 58.443 g

If sample was 100% MgCl2 molarity = 3 x (0.5 / 95.21 ) = 0.01575 Mol/liter of particles

If sample was 100% NaCl molarity = 2 x (0.5 / 58.55 ) = 0.01711 Mol/liter of particles

Let x = Percent MgCl2

0.01575 x + 0.01711 x (1-x) = 0.01631

0.01575 x +0.01711 - 0.01711 x = 0.01631

0.0008 = 0.00136 x

x = 0.588

x % = MgCl2 = 58.8%

MgCl2 mass percent = 58.8 %