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Lisa [10]
4 years ago
6

A 0.5922 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 1.3487 g. What is the mass percentage of chlorine in the original compound?
Chemistry
1 answer:
Elanso [62]4 years ago
6 0

<u>Answer:</u> The mass percent chlorine in pure chloride compound is 56.31 %.

<u>Explanation:</u>

We are given:

Mass of AgCl precipitated = 1.3487 g

We know that:

Molar mass of AgCl = 143.32 g/mol

Molar mass of Chlorine atom = 35.45 g/mol

As, all the chlorine in the sample is precipitated to silver chloride. So, the mass of chlorine in silver chloride will be equal to the mass of chlorine present in the sample.

To calculate the mass of chlorine in given mass of AgCl, we apply unitary method:

In 143.32 g of silver chloride, mass of chlorine present is 35.45 g

So, in 1.3487 g of silver chloride, mass of chlorine present will be = \frac{35.45g}{143.32g}\times 1.3487g=0.3335g

To calculate the percentage composition of chlorine in pure chloride compound, we use the equation:

\%\text{ composition of chlorine}=\frac{\text{Mass of chlorine}}{\text{Mass of pure chloride compound}}\times 100

Mass of pure chloride compound = 0.5922 g

Mass of chlorine = 0.3335 g

Putting values in above equation, we get:

\%\text{ composition of chlorine}=\frac{0.3335g}{0.5922g}\times 100=56.31\%

Hence, the mass percent chlorine in pure chloride compound is 56.31 %.

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