Question

Dr. Suzanne acquires a quantity of a 923ppm sodium chloride solution (solution a) Dr Suzanne takes a small volume of solution A and adds a bunch of water to it. This dilution solution (Solution B) has a concentration of 175 ppm. Dr Suzanne forgot what volume she used, so she makes a plan to find it, she reacted all of Solution B with excess Lead II Nitrate solution, producing 0.399g of precipitate. Using the information The volume of Solution A used to prepare solution B is?

Answer 1

Pb(NO₃)_2(aq) + 2NaCl_(aq) -->2 NaNO_3(aq) + PbCl_2(s);

m(NaCl)=m(PbCl2)*2Mr(NaCl)/Mr(PbCl2)=(0.399*2*58)/278=0.1665 g=166.5 mg;

V(solution A)=166.5/923=0.18 L=180 mL;

Answer: 180 ml of soltion A was used;