An unknown liquid inside a metal sylinder contain 0.84 grams of Lead(II) Chloride (PbCl2) per 100 ml of soluti

Question

An unknown liquid inside a metal sylinder contain 0.84 grams of Lead(II)Chloride (PbCl2) per 100 ml of solution. what is its molarity?

Answer 1

Atomic weights: Pb=207 Cl=35.5 PbCl2=278

Let the solution be called S

0.84gPbCl2/100mLS x 1molPbCl2/278gPbCl2 x 1000mLS/1LS = (0.84)(1000)/(100)(278) = 0.030mol/L = 0.030M

The 0.84g PbCl2 and the 100 mL solution are given. The next factor comes from the formulas weight of PbCl2. The g PbCl2 cancel, leaving the mols PbCl2. The next factor comes from the equivalency of 1000 mL per L. The mL solution cancel, leaving L solution.