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Difficult A level chem question

The Question:

“The following experiment was carried out to find the formula for an ionic chloride, MCln, containing M^n+ and Cl^- ions. 1.21g of the chloride was dissolved in water and made up to 250cm^3 of solution. 25.0cm^3 of this solution was titrated with silver nitrate solution containing 0.0500 mol dm^-3 of Ag+ (aq), using potassium chromate (VII) solution as an indicator. 20.0cm^3 of silver nitrate solution was required to reach the end-point.

Ag^+ (aq) + Cl^- (aq) —> AgCl (s)

a) Calculate the number of moles of silver ions used in the titration.
b) Calculate the mass of Cl^- and hence of M in the chloride.
c) Find the empirical formula of the chloride.”

I could only complete a) which I found to be 0.001 mol.

Regarding b) and c), it’s perplexingly discombobulating. I had a look at the answer as I practically gave up, and the mass of Cl- was 0.355g, and the mass for M was 0.855g, but I don’t know how to get there. Any help would be appreciated.
Original post by Windup
The Question:
“The following experiment was carried out to find the formula for an ionic chloride, MCln, containing M^n+ and Cl^- ions. 1.21g of the chloride was dissolved in water and made up to 250cm^3 of solution. 25.0cm^3 of this solution was titrated with silver nitrate solution containing 0.0500 mol dm^-3 of Ag+ (aq), using potassium chromate (VII) solution as an indicator. 20.0cm^3 of silver nitrate solution was required to reach the end-point.
Ag^+ (aq) + Cl^- (aq) —> AgCl (s)
a) Calculate the number of moles of silver ions used in the titration.
b) Calculate the mass of Cl^- and hence of M in the chloride.
c) Find the empirical formula of the chloride.”
I could only complete a) which I found to be 0.001 mol.
Regarding b) and c), it’s perplexingly discombobulating. I had a look at the answer as I practically gave up, and the mass of Cl- was 0.355g, and the mass for M was 0.855g, but I don’t know how to get there. Any help would be appreciated.

For b)
Ag+ : Cl- is 1:1, so moles of Cl- in 25cm^3 is 0.001 mol.
Moles of Cl- in 250cm3 = 0.001 x 250/25 = 0.01 mol
Mass = moles X Mr. Mass chloride ion = 0.01 x 35.5 = 0.355g
Mass salt (1.21g) - mass chloride (0.355g) = mass of M, 0.855g

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