Question

Weight of copper oxide obtained by treating 2.16g metallic copper with Nitric acid and subsequent ignition was 2.7g. in another experiment 1.15g of CuO on reaction gives 0.92g Cu. Show that the result illustrate law of constant proportion.​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$


$\underline{ \bf{Law \: of \: Constant \: Proportion :}}$

According to this law, a compound may be obtain from different sources but the ratio of each component (by weight) remains same.


Now,


$\underline{ \bf{EXPERIMENT \: \: \: 1 :}}$

Mass of CuO = 2.7 g

Mass of Cu = 2.16 g

So, Mass of O = (2.7 - 2.16) g

= 0.54 g


Now,

Cu : O = 2.16 : 0.54

= $\frac{2.16}{0.54}$ : $\frac{0.54}{0.54}$

= 4 : 1


$\underline{ \bf{EXPERIMENT \: \: \: 2 :}}$

Mass of CuO = 1.15 g

Mass of Cu = 0.92 g

Mass of O = (1.15 - 0.92) g

= 0.23 g


Now,

Cu : O = 0.92 : 0.23

= $\frac{0.92}{0.23}$ : $\frac{0.23}{0.23}$

= 4 : 1


So, here we can see in both experiment we get the ratio of Cu and O as same i.e 4 : 1.

Therefore, we can say that this illustrate law of Constant Proportion.

Answer 2

ANSWER :

CASE 1 :

CuO = 2.7 g (Given)

Cu = 2.16 g (Given)

So, Mass of O = 2.7g - 2.16g

= 0.54 g

Now, Ratio

Cu : O = 2.16 : 0.54

= 2.16/0.54 : 0.54/0.54

= 4 : 1

CASE II :

CuO = 1.15 g

Cu = 0.92 g

Mass of O = 1.15g - 0.92g

= 0.23 g

Now, Ratio

Cu : O = 0.92 : 0.23

= 0.92/0.23 : 0.23/0.23

= 4 : 1

So, here we can see in both experiment we get the ratio of Cu and O as same i.e 4 : 1.

Hence Law of Constant Proportion is verified