1. The equation for the solution process is
MX3 (s) M3+ (aq) + 3X�?/SUP> (aq)
Equilibrium: �?nbsp; x 3x
Therefore, Ksp = [M3+][X�?/SUP>]3, 8.47 X 10�?5 = (x)(3x)3 = 27X4, x = 1.33 X 10�? M
So, the molar solubility is 1.33 X 10�? M, and the concentration of X�?/SUP> is 3x.
3(1.33 X 10�? M) = 3.99 X 10�? M or 0.399 mM.
2. Looks OK, except it should be �?.48 kJ/mol (negative).
3. I always do these molecular weight calculations from a colligative property in three steps. In this case, the colligative property is osmotic pressure, Π.
a) Calculate the concentration (molarity here).
Π = MRT, M = Π / RT = 0.044 atm / (0.08206 L atm/mol K)(298.15 K) = 0.0017984 M.
b) From the concentration and volume, calculate the moles of solute.
Molarity = moles / L, moles = molarity X L = (0.0017984 mol/L)(0.500 L) = 0.0008992 mol
c) Divide the grams of solute by the moles of solute to get the molar mass in units of g/mol.
8.9 g / 0.0008992 mol = 9898 g/mol
4. To convert molarity to molality, moles / kg of solvent, we can start by assuming that we have 1 liter of the solution. This 1 liter of solution will then have 14.1 moles of CH3OH in it. Convert 14.1 moles of CH3OH to grams.
Next, calculate the total mass of the 1 liter of solution: mass = volume X density:
mass = 1000 mL X 0.859 g/mL = 859 g
Now get the mass of the solvent alone by subtracting the mass of the CH3OH from the total mass of the solution:
mass of H2O = 859 g �?nbsp; grams of CH3OH
Convert grams of H2O to kg, and then plug the numbers into the molality formula:
molality = moles of CH3OH / kg of H2O
That should do it!
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