Question

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)?

• Your answer should have three significant figures (round to the nearest hundredth).

Answer 1

Answer

1.06 mol/L

Explanation

The balance equation is

MgCl2 + Pb(NO3)2 --------> PbCl2 + Mg(NO3)2

From the balanced equation we know that 1mole of PbCl2 comes from 1mole of MgCl2

Number of moles = mass/molar mass

Number of moles of PbCl2 = 7.35g/278.1g/mol = 0.02643mol

0.02643 moles of PbCl2 comes from 0.02643moles of MgCl2

MgCl2 contains 2 Cl^-

Number of moles Cl^- = 2× 0.02643mol = 0.05286mol

Molar concentration of Cl^- = (0.05286mol/50ml)×1000ml= 1.06 mol/L