Question

A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 12.11 g CaCl2 ( s ) is obtained from 200.0 mL of the original solution.

Answer 1

Answer:

0.218 M of Pb(NO3)2

Explanation:

Equation of the reaction

Pb(NO3)2(aq) + 2NaCl(aq) --> PbCl2(s) + 2NaNO3(aq)

1 mole of Pb(NO3)2 reacts to precipitate 1 mole of PbCl2

Molar mass of PbCl2 = 207 + (35.5*2)

= 278 g/mol

Number of moles of PbCl2 precipitated = mass/molar mass

= 12.11/278

= 0.04356 mol

Since 0.04356 moles of PbCl2 was precipitated, therefore by stoichiometry; 0.04356 moles of Pb(NO3)2 reacted.

Molarity is defined as the number of moles of solute in 1 liter of solution.

Molarity = number of moles/volumes

= 0.04356/0.2

= 0.218 M