Question

Ca(s) reacts with water forming calcium hydroxide. If a 3.00 g chunk of calcium is reacted with water and the solution is further diluted with water to a volume of 250 mL at 60.0 °C (d = 0.981 g/mL) calculate the following: a) the concentration of calcium hydroxide b) the concentration of hydroxide anions c) the molality of calcium hydroxide d) the molality of hydroxide anions e) the mole fraction of calcium hydroxide f) the mole fraction of hydroxide anions g) the mass% of calcium hydroxide h) the mass% of hydroxide anions

Answer 1

Sol:-

(a)

..........Ca(s).........+............2H₂O (l) -----------> Ca(OH)₂ (aq) ............+...........H₂ (g)

.........1mol.........................2mol......................1 mol........................................1mol

..........40g/mol...............2 x 18g/mol................74g/mol..................................2g/mol

we know

40 g of Ca gives = 74 g of Ca(OH)₂

so

3.00 g of Ca gives = 74 g x 3.00 g / 40 g of Ca(OH)₂ = 5.55 g of Ca(OH)₂

Now mass of Ca(OH)₂ = 5.55 g

therefore

Number of moles of Ca(OH)₂ = given mass of Ca(OH)₂ / gram molar mass of Ca(OH)₂

Number of moles of Ca(OH)₂ = 5.55g / 74 g/mol = 0.075 mol

Concentration (i.e Molarity ) = number of moles / volume of the solution in L

Concentration (i.e Molarity ) = 0.075 mol / 0.250 L = 0.3 mol/L = 0.3 M

Hence concentration of calcium hydroxide = 0.3 M

(b)

........... Ca(OH)₂ (aq) -------> Ca²⁺(aq)..............+.............2OH^-(aq)

............0.3 M.........................0.3 M..............................2x0.3M = 0.6 M

So Concentration of  hydroxide anions i.e OH^- = 0.6 M

(c)

0.3 M Ca(OH)₂ means 0.3 moles of Ca(OH)₂ present in 1000 mL of solution.

so

Number of moles of Ca(OH)₂ = 0.3 mol

and

Mass of Ca(OH)₂ (solute) = moles x gram molar mass of Ca(OH)₂

Mass of Ca(OH)₂ (solute) = 0.3 mol x 74g/mol = 22.2 g

also Volume of the solution = 1000 mL

given density of the solution = 0.981 g/mL

therefore

Mass of solution = density x volume = 0.981 g/mL x 1000 mL = 981 g

we know

Mass of solution = mass of solute i.e Ca(OH)₂ + Mass of solvent i.e H₂O

and

Mass of solvent i.e H₂O = Mass of solution - mass of solute i.e Ca(OH)₂

Mass of solvent i.e H₂O = 981 g - 22.2 g = 958.8 g = 0.9588 kg

Molality = Number of moles of solute / mass of solvent in kg

Molality of Ca(OH)₂ = 0.3 mol / 0.9588 kg = 0.313 mol/kg

Hence molality of calcium hydroxide = 0.313 mol/kg

(d)

........... Ca(OH)₂ (aq) -------> Ca²⁺(aq)..............+.............2OH^-(aq)

............0.313 mol/kg............0.313mol/kg......................2x0.313mol/kg = 0.626 mol/kg

Hence Molality of hydroxide anions = 0.626 mol/kg