Question

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 13.93 g of PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

? M

Answer 1

Molar mass of PbCl2 = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

mass of PbCl2 = 13.93 g

we have below equation to be used:

number of mol of PbCl2,

n = mass of PbCl2/molar mass of PbCl2

=(13.93 g)/(278.1 g/mol)

= 5.009*10^-2 mol

the reaction that took place is:

Pb(NO3)2 + 2 NaCl —> PbCl2 + 2 NaNO3

from reaction,

mol of Pb(NO3)2 reacted = number of mol of PbCl2

= 5.009*10^-2 mol

volume , V = 200 mL= 0.200 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 5.009*10^-2/0.2

= 0.250 M

Answer: 0.250 M