Oftentimes solubility of a compound limits the concentration of the solution that can be prepared. Use the solubility data given with each compound shown below to determine which compound would allow the preparation of a 10.0 Molar solution.

A) AgNO3 (solubility = 122 g/100 g H2O)

B) KCl (solubility = 34.0 g/100 g H2O)

C) NaNO3 (solubility = 89.0 g/100 g H2O)

D) NH4Cl (solubility = 41.1 g/100 g H2O)

E) none of the above

NaNO3 (solubility = 89.0 g/100 g H2O)

Explanation:

The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.

Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)

Molar mass of NaNO3 = 23+14+3 (16) = 85gmol-1

Mass of solute=89.0g

Amount of solute = mass of NaNO3/molar mass of NaNO3

Amount of solute = 89.0g/85.0 gmol-1

= 1.0moles of NaNO3

Note that 100g of water=100cm^3 of water.

If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,

x moles of NaNO3 will dissolve in 1000cm^3 of water

x = 1.0 * 1000 / 100

x = 10.0 moles of NaNO3