You have two 466.0 ml aqueous solutions. solution a is a solution of silver nitrate, and solution b is a solution of potassium chromate. the masses of the solutes in each of the solutions are the same. when the solutions are added together, a blood-red precipitate forms. after the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g. (a) calculate the concentration of the potassium ions in the original potassium chromate solution.(b) calculate the concentration of the chromate ions in the final solution

The concentration of the potassium ions in the original potassium chromate solution is 4.2927 mol/L.

The concentration of the chromate ions in the final solution is 1.0731 mol/L.

Explanation:

Volume of solution A i.e. solution of silver nitrate = 466.0 mL = 0.466 L

Volume of solution B i.e. solution of potassium chromate = 466.0 mL = 0.466 L

Moles of silver chromate =

According to reaction , 1 mol of silver chromate is produce from 2 moles of silver nitrate.

Then, 1.0002 moles of silver chromate will be formed from:

of silver nitrate.

According to reaction , 1 mol of silver chromate is produce from 1 mole of potassium chromate.

Then, 1.0002 moles of silver chromate will be formed from:

of potassium chromate

a) The concentration of the potassium ions in the original potassium chromate solution.

Volume of the original solution = 0.466 L

1 mol of potassium chromate dissociates into 2mol of potassium ions and 1 mol of chromate ions:

Moles of potassium ions = 2 × 1.0002 mol = 2.0004 mol

b) The concentration of the chromate ions in the final solution

Volume of the final solution = 0.466 L + 0.466 L

Moles of chromate ions = 1 × 1.0002 mol = 1.0002 mol

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation: