posted by Katherine on .
Suppose we have a solution of lead nitrate, Pb(NO3). 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 10.27g of PbCl2(s) is obtained from 200.0mL of the original solution.
Calculate the molarity of Pb(NO3)2(aq)solution.
10.27g PbCl2/ molar mass PbCl2 = ? mols PbCl2.
That means that many mols Pb(NO3)2. That was in 200 mL solution;M = mols/L.