# In two separate experiments 0.36g and 0.71g of chlorine combined with…

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## In two separate experiments 0.36g and 0.71g of chlorine combined with a metal X to give Y and Z, an analysis showed that Y and Z contain 0.20g and 0.40g of X respectively. The data above represents the law of

A. multiple proportion

B. conservation of mass

C. constant composition

D. reciprocal proportion.

C

## SOLUTION…

Let Y = XCl

If the mass of chlorine in Y is 0.36g,

And the mass of chlorine in Z is 0.71g, it means chlorine has doubled.

Also,

If the mass of X in Y is 0.20g,

And the mass of X in Z is 0.40g, it means the number of X has also doubled.

Therefore,

Z = 2(Y) = 2(XCl)

This demonstrates the law of constant composition which states that when elements combine, they do so in simple whole number ratio.

To put it in simple terms, when the mass of an element in a compound doubles, the law of constant composition states that other fellow elements in the compound will also have a doubled mass.

To explain it further in more simpler terms, in the first experiment, 0.36g of chlorine combined with 0.20g of X to give Y.

In the second experiment, 0.71g of chlorine, which is double that of the 1st experiment, combined with 0.40g of X, which is also double.

So, basically, we have the same substance, but at different quantities. if the first compound is XCl, the second will be 2(XCl). This is constant composition.

### Now for the right answer to the above question:

1. Option A is incorrect.
2. Option B is incorrect.
3. C is correct.
4. D is not the correct answer.

## KEY-POINTS…

• The law of constant composition explains that an increase in mass of one of the elements in a compound implies an increase by the same factor for other fellow elements in the compound.
• For example, X2Y3 = X4Y6 = X6Y9 ….

Use the questions and answers session to raise further questions and deal more on this topic…

/ culled from 2017 JAMB-UTME Chemistry question 4 /

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