Question

1.20 g of MgCO3 are dissolved in water and brought to a final volume of 700 mL. What is the molar concentration of the solution?

2.Calculate the normality of the solution.

3. If the MgCO3 completely dissociates to Mg2+ and CO3 2- , what are the concentrations of Mg2+ and CO3 2- (in g/L) in the solution?

4. Calculate the mass fraction of Mg2+ in the solution.

5. Calculate the molar fraction of C in solution.

Answer #1

Molar mass of MgCO3 is 84.3 g/mol

Number of moles of MgCO3 = mass/molar mass = 1.2/84.3 = 0.0142 moles

Final volume = 700 mL

since Molarity is the concentration in 1000 mL

on converting it to 1000 mL

Concentration = (0.0142/700) 1000 = 0.020 M

2. equivalent weight of MgCO3 = 42.15 g /mole

Let X = molar mass/ equivalent mass

the Normality = molarity x X

Here X =2

the Normality = 2 x 0.02 = 0.04 N

3. Molar mass of Mg = 24.3 g/mol

Thus mass of Mg in 1.2 g = 1.2 x 24.3/84.3 = 0.345 g in 700 mL

concentration in g/mL = 4.94 x 10^{-4} g

4. total weight of solution = 700 + 1.2 = 701.2g (density of water is 1 g/mL)

Mass fraction of Mg = mass of Mg/total mass of the solution =
0.345 /701.2 = 4.92 x 10^{-4} g

5. Moles of water = 700/18 = 38.88 moles

Moles of C = moles of MgCO3

molar fraction = moles of C /total moles ) = 0.0142/38.90 = 3.65
x 10^{-4}

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