Asolution of nacl(aq) is added slowly to a solution of lead nitrate, pb(no3)2(aq) , until no further precipitation occurs. the precipitate is collected by filtration, dried, and weighed. a total of 18.86 g pbcl2(s) is obtained from 200.0 ml of the original solution. calculate the molarity of the pb(no3)2(aq) solution.

The molarity of solution is 0.34 M.

Explanation:

For lead chloride:

Given mass of lead chloride = 18.86 g

Molar mass of lead chloride = 278.1 g/mol

Putting values in above equation, we get:

By Stoichiometry of the reaction:

1 mole of lead chloride is formed by 1 mole of lead nitrate

So, 0.068 moles of lead chloride will be formed from = of lead nitrate

We are given:

Volume of solution = 200 mL = 0.200 L   (Conversion factor: 1 L = 1000 mL)

Moles of lead nitrate = 0.068 moles

Putting values in above equation, we get:

Hence, the molarity of solution is 0.34 M.

the answer is symmetrical : )