Question

M+ is an unknown metal cation with a +1 charge. A student dissolves the chloride of the unknown metal, MCl, in enough water to make 100.0 mL of solution. The student then mixes the solution with excess AgNO3 solution, causing AgCl to precipitate. The student collects the precipitate by filtration, dries, it and records the data shown below. (The molar mass of AgCl is 143 g/mol.) Mass of unknown chloride, MCl Mass of filter paper Mass of filter paper plus AgCl ppt 0.42 g 1.00 g 2.43 g Find the identity of metal chloride.
(A) lithium chloride
(B) sodium chloride
(C) magnesium chloride

Answer 1

Mass of AgCl = (mass of filter paper plus AgCl) - (mass of filter paper)
Mass of AgCl = 2.43 g - 1.00 g = 1.43 g AgCl 

1.43 g AgCl (1 mole AgCl / 143 g AgCl) = 0.010 moles Ag left

Since there is 1 is to 1 ratio between the ions and the compound, then moles Ag+ = moles AgCl = 0.010 and moles Cl- = moles AgCl = 0.010. 

The formula of the unknown chloride is MCl, so:

0.010 moles Cl- (1 mole M+ / 1 mole Cl-) = 0.010 moles M+ 

0.010 moles Cl- (35.5 g Cl- / 1 mole Cl-) = 0.37 g Cl- 

Mass M+ = Mass of MCl - mass of Cl- = 0.42 g - 0.37 g = 0.05 g. 

0.05 g M+ / 0.010 moles M+ = 5 g/mole

I think the best answer would be option A. The metal chloride must be lithium chloride since the atomic mass of lithium is close to the calculated mass above.

Answer 2

Looks like you're missing an answer choice. I have this question on my practice test for AP Chemistry, the correct answer would be KCl.

mass AgCl = 2.23 - 0.80 = 1.43g AgCl

1.43g AgCl / 143g/mol AgCl = 0.01 moles AgCl

- AgCl has a mole ratio of 1:1 so:

     moles Ag+ = moles Cl-, therefore Ag+ has 0.01 moles and Cl- has 0.01 moles

- MCl also has a 1:1 ration

      moles Cl- = moles M+

      - 0.01 moles of M+ and Cl-

0.01 Cl- = x/35.45 = 0.3545g Cl-

mass MCl = 0.74g

0.74g MCl - 0.3545g Cl- = 0.3955g M+

0.3955g M+/x =0.010 mol M+

x= 39.55g M+

K+ has a molar mass of approximately 39.10

Therefore KCl is your answer