Question

A 1.63-g sample of a metal chloride, MCI 2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed
weighed 3.48 g. Calculate the molar mass of M.
a. 72.4 g/mol
b.67 g/mol
c. 64 g/mol
d. 32 g/mol
e. 70.9 g/mol

Answer 1

Answer:

The molar mass of M is 44.06 g/mol

Explanation:

The reaction between a metal chloride and silver nitrate will be given by the balanced equation:

MCl(aq)+ AgNO₃(aq) →MNO₃(aq) + AgCl(s)

We are required to calculate the molar mass of M

We are going to use the following steps:

Step 1: Number of moles of silver nitrate

Number of moles = Mass of a compound ÷ Molar mass of the compound

Molar mass of silver nitrate is 169.87 g/mol

Therefore;

Number of Moles = 3.48 g ÷ 169.87 g/mol

                             = 0.0205 moles

Step 2: Moles of MCl that reacted

From the equation; 1 mole of MCl reacts with 1 mole of silver nitrate

Therefore; the mole ratio of MCl : AgNO3 is 1: 1

Thus;

Moles of MCl = 0.0205 moles

Step 3: Molar mass of MCl

From the question 1.63 g sample of MCl was used

Therefore;

1.63 g = 0.0205 moles

Molar mass is the relative formula mass of one mole of a compound

Therefore;

Molar mass = 1.63 g ÷ 0.0205 moles

                   = 79.51 g/mol

Step 3: Molar mass of M

The relative formula mass of MCl is 79.51 g/mol

Atomic mass of Cl = 35.45

Thus;

79.51 g/mol = M + 35.45 g

 M = 79.51 - 35.45

     = 44.06

Thus, the molar mass of M is 44.06 g/mol