A 0.4066 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 0.9260 g. What is the mass percentage of chlorine in the original compound? %

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Abdulazeez10 Abdulazeez10 · Answer 1 · 2020-09-09T05:19:09+02:00

Answer:

The mass percentage of chlorine in the original compound is 56.32%

Explanation:

From the question,

Mass of original compound = 0.4066 g

To, determine the mass percentage of chlorine in the original compound,

First, we will determine the mass of chlorine present in the precipitated AgCl

Mass of precipitated AgCl = 0.9260 g

Molar mass of AgCl

Ag = 107.87 g/mol; Cl = 35.45 g/mol

Then, molar mass of AgCl = 107.87 + 35.45

= 143.32 g/mol

Now,

If 35.45 g of chlorine is present in 143.32 g of AgCl

Then, [tex]x[/tex] g of chlorine will be present in 0.9260 g of AgCl

[tex]x = \frac{35.45 \times 0.9260}{143.32}[/tex]

[tex]x = 0.2290[/tex] g

Hence, 0.2290 g of chlorine is present in 0.9260 g of AgCl.

Since all of the chloride ion is precipitated,

then 0.2290 g of chlorine is present in the original chloride compound.

Now, for the mass percentage of chlorine in the original compound,

mass percentage of chlorine =

(mass of chlorine in the compound /mass of the chloride compound) × 100%

mass of chlorine in the compound = 0.2290 g

mass of the chloride compound = 0.4066 g

∴ mass percentage of chlorine = [tex]\frac{0.2290}{0.4066}[/tex] × 100%

mass percentage of chlorine = 56.32%

Hence, the mass percentage of chlorine in the original compound is 56.32%