Question

Copper oxide was prepared by 2 different methods. In one case, 1.75g of the metal gave 2.19g of oxide. In the second case, 1.14g of the metal gave 1.43g of the oxide. Show that the given data illustrate the law of constant proportions.

Answer 1

$\text{\underline{\underline{In\:case\:I:}}}$

We know that,

Mass of copper = 1.75 g

Mass of copper oxide = 2.19 g


So,

$\sf{\% \: of \: copper \: in \: the \: oxide = \frac{Mass \:of \: copper }{Mass \: of \: copper \: oxide} \times 100}$

$\sf{ = \frac{1.75}{2.19} \times 100}$

= $\sf\boxed{ 79.9\%}$


$\text{\underline{Therefore,}}$

$\sf{\% \: of \: oxygen = 100 - 79.9 =\boxed {20.1\%}}$


$\text{\underline{\underline{In\:case\:II:}}}$

We know that,

Mass of copper = 1.14 g

Mass of copper oxide = 1.43 g


So,

$\sf{\% \: of \: copper \: in \: the \: oxide = \frac{Mass \:of \: copper }{Mass \: of \: copper \: oxide} \times 100}$

$\sf{ = \frac{1.14}{1.43} \times 100}$

= $\sf\boxed{ 79.7\%}$


$\text{\underline{Therefore,}}$

$\sf{\% \: of \: oxygen = 100 - 79.7 = \boxed{20.3\%}}$


$\text{\underline{Thus,}}$

Copper oxide prepared by any of the given methods contain copper and oxygen in the same proportion by mass.


$\text{\underline{Hence,}}$

It proves the law of constant proportions.

Answer 2

Step 1: In the first experiment.

2.19 g of copper oxide contained 1.75 g of Cu.

100 g of copper oxide contained

Step 2: In the second experiment.

1.43 g of copper oxide contained 1.14 g of copper.

100 g of copper oxide contained

Step 3: In the third experiment.

1.83 g of copper oxide contained 1.46 g of copper

100 g of copper oxide contained

Thus the percentage of copper in copper oxide derived from all the three experiments is nearly the same. Hence, the above data illustrate the law of constant composition.

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