Question

1) When 18.0 mL of a 5.23×10-4 M sodium carbonate solution is combined with 15.0 mL of a 4.47×10-4 M cobalt(II) sulfate solution does a precipitate form? (yes or no) For these conditions the Reaction Quotient, Q, is equal to

2) The freezing point of water, H₂O, is 0.000 °C at 1 atmosphere. K_f(water) = 1.86 °C/m
In a laboratory experiment, students synthesized a new compound and found that when 11.98 grams of the compound were dissolved in 296.5 grams of water, the solution began to freeze at -1.250 °C. The compound was also found to be nonvolatile and a non-electrolyte.
What is the molecular weight they determined for this compound?
__g/mol

3) In a laboratory experiment, students synthesized a new compound and found that when 14.01 grams of the compound were dissolved to make 265.8 mL of a ethanol solution, the osmotic pressure generated was 1.44 atm at 298 K. The compound was also found to be nonvolatile and a non-electrolyte.
What is the molecular weight they determined for this compound?
__ g/mol

Answer 1

1) Final volume of the solutions = (18.0 + 15.0) mL = 33.0 mL. Now, we have to calculate the concentration of Co²⁺ and CO₃^- in final solution, because precipitation, if occurs, will be of CoCO₃, using condition of equivalency, which is:

, where, V indicates volume and S indicates concentration.

For sodium carbonate, V_initial=18.0 mL, S_initial=5.23×10^-4 M, V_final = 33.0 mL,

so,

and that is the final concentration of carbonate.

For cobalt(II) sulphate, V_initial=15.0 mL, S_initial=4.47×10^-4 M, V_final = 33.0 mL,

so,

and that is the final concentration of Co(II)

So, reaction quocient = Q = [Co²⁺] [CO₃^-] = 2.032 x 10^-4 x 2.853 x 10^-4 = 5.797 x 10^-8.

K_sp of CoCO₃ is 1.0 x 10^-12, so Q > K_sp, thus the precipitation will occur.

2) The depression of freezing point has the equation, , where m₂ is the molal concentration, i.e, no of moles of solute dissolved in 1000 g of solvent.

Here, ,

So, m₂ = 1.250 ^oC / (1.86 ^oC/m) = 0.672 m, i.e, 0.672 moles in 1000 g solvent or water.

Now, in 296.5 g of water solute dissolved is 11.98 g,

So, in 1000 g solute dissolved is (11.98 x 1000 / 296.5) g = 40.405 g

So, 0.672 moles = 40.405 g

Thus, 1 mol = 60.126 g

Thus, molecular weight is 60.126 g.mol^-1.

3) Osmotic pressure has the equation: , where, is osmotic pressure, C is molar concentratoion and T is Kelvin temperature.

Here, = 1.44 atm, R = 0.0821 L.atm.K^-1.mol^-1, T = 298 K,

So, C = (1.44 atm) / (0.0821 L.atm.K^-1.mol^-1 x 298 K) = 0.0588 mol.L^-1

Now, for our solution,

265.8 mL solution contains 14.01 g compound

So, 1000 mL .i.e, 1 L contains (14.01 x 1000 / 265.8) g = 52.709 g

So, 0.0588 mol = 52.709 g

thus, 1 mol = (52.709 / 0.0588) g = 896.41 g

So. molecular weight = 896.41 g.mol^-1.