Question

I mixed 10g sodium carbonate in 48 cold water and 14 grams of calcium hypochlorite whats the final percentage of the solution?

Answer 1

Consider what happens at a molecular level.

When you deposit each solute, it will be solvated by the water, dissociated into its ions.

If you're interested in the maximum concentration of `NaClO` in your solution (even though this is dissociated),

We have,

`10"g" * (Na_2CO_3)/(106"g") * (2Na^+)/(Na_2CO_3) approx 0.189"mol"` of sodium ions, and,

`14"g" * (Ca(ClO)_2)/(143"g") * (2ClO^(-))/(Ca(ClO)_2) approx 0.196"mol"` of hypochlorite ions.

The "limiting" ion in this case is sodium, because there is less available than hypochlorite given your data. If we wanted to isolate this salt by some technique, then we could only isolate approximately `0.189"mol"` worth at best.

Hence, the "concentration" of this new salt you want would be,

`(0.189"mol")/(48"mL" * ("L")/(10^3"mL")) approx 39.4"M"`

given your data.

Note: this concentration is grossly unreasonable, because that much inorganic salt couldn't foreseeably dissolve in only that volume of water. The given concentration is theoretical, and in the lab the solution would probably be supersaturated.