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 17.19 g of pbcl2(s) is obtained from 200.0 ml of the original solution. calculate the molarity of the pb(no3)2(aq) solution.

The solution to calculate the molarity of the Pb(NO3)2(aq) is below; 

(17.19 g PbCl2) / (278.1164 g PbCl2/mol) x (1 mol Pb(NO3)2 / 1 mol PbCl2) / (0.2000 L) 

Be careful on inserting the data on the calculator when you are solving it may give you an inaccurate answer.

= 0.3090 mol Pb(NO3)2/L
The molarity of the pb(no3)2(aq is 0.3090 mol Pb(NO3)2/L

I’m not sure what your trying to say here.. can you clarify again?