A12.00g sample of addition of an excess of lead (ii) nitrate to a 50.0ml solution of magnesium chloride caused a formation of 7.35g of lead (ii) chloride precipitate. what was the molar concentration of chloride ions in the solution (in mol/l)?

The concentration of chloride ions in the solution is 1.056 mol/L

Explanation:

The chemical equation for the reaction of lead (II) nitrate and magnesium chloride follows:

All the chloride ions are getting converted to lead (II) chloride

To calculate the number of moles, we use the equation:

Given mass of lead (II) chloride = 7.35 g

Molar mass of lead (II) chloride = 278.1 g/mol

Putting values in above equation, we get:

1 mole of lead (II) chloride produces 1 mole of lead ions and 2 moles of chloride ions.

So, moles of chloride ions = (2 × 0.0264) = 0.0528 moles

To calculate the molarity of solution, we use the equation:

Moles of chloride ions = 0.0528 moles

Volume of solution = 50.0 mL

Putting values in above equation, we get:

Hence, the concentration of chloride ions in the solution is 1.056 mol/L

particle   p is an electron and particle q is a neutron