What is the molar concentration of chloride ions in a
solutionprepared by mixing 100ml of 2.0 m kcl with 50 l of a 1.5
mcacl2 solution?

The concentration of chloride ions in the final solution is 3 M.

Explanation:

The number of moles present in a solution can be calculated as follows:

number of moles = concentration in molarity * volume

In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻

For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:

number of moles of Cl⁻ = 2 * number of moles of  CaCl₂

number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻

The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.

Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:

Concentration = number of moles of Cl⁻ / volume

Concentration = 150.2 mol / 50.1 l = 3.0 M

b is the correct