Question

water was added to 100g of a salt mgcl2 to produce 80.0cm3 of saturated solution at 25°c. its solubility at 25°c is 10 mol/dm 3. calculate the mass of the salt which remains undissolved

Answer 1

Solubility refers to the mass of a solute per 100 g of a solvent .

can be express as g/100 g of water.

converting mass of salt into mole.

1 mole of cacl₂ = 40 + 2(35.5) = 121g

$100g * (1mol/121g) = 0.826 mol$

solution of into dm³

$80cm^3 *(1dm^3/1000cm^3) = 0.08dm^3$

solubility = 0.826mol /0.08dm³ =10.325 mol/dm³

subtracting the actual solubility with that of the salt added

at the 25^oc 10.325 mol/dm³ - 10.0mol/dm³ = 0.325mol

0.325 mol is equivalent to

$0.325mol* (121g/1mol) = 39.325 g$ this is the mass that remained un dissolved