The question is incomplete; the complete question is;
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 0.74 g
Mass of filter paper 0.80 g
Mass of filter paper plus AgCl precipitate 2.23 g
1. What is the identity of the metal chloride?
(A) NaCl
(B) KCl
(C) CuCl
(D) LiCl
Answer:
KCl
Explanation:
Mass of paper= 0.80 g
Mass of paper and AgCl=2.23
Hence mass AgCl = 2.23 - 0.80 = 1.43g AgCl
Number of moles of AgCl= 1.43g AgCl / 143g/mol AgCl = 0.01 moles AgCl
Since Ag+ and Cl- are in a mole ratio of 1:1 so:
number moles Ag+ = number moles Cl-,
Thus Ag+ has 0.01 moles and Cl- has 0.01 moles
- MCl is also in a 1:1 ratio
Hence;
number of moles of moles Cl- =number of moles of moles M+ = 0.01 moles of M+ and Cl-
Thus;
0.01 Cl- = x/35.45 = 0.3545g of Cl-
mass of MCl = 0.74g
Mass of M+ =0.74g MCl - 0.3545g Cl- = 0.3955g of M+
Number of moles of M+ =0.3955g M+/x =0.010 mol M+
x= 39.55g of M+
K+ has a molar mass of approximately 39.10
Hence KCl is the answer
