a metal forms two oxides. The higher oxide contains 80% metal. 0.72g of the lower oxide gave 0.8g of the higher oxide when oxidised. show that the data illustrate the law of multiple proportions.​

0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.

Solution given:

For higher oxide:

contains 80% metal

weight =80% 0f 0.8g=0.64g.

weight of Oxygen=0.8-0.6=0.16g

For lower oxide:

the metal's weight is fixed.

weight of oxygen=0.72g-0.64g=0.08g

Now

Ratio of oxygen combined with fixed amount of metal is :

0.16:0.08

=2:1

since it is a simple ratio .

so;

It follows the law of multiple proportions.

it is a good conductor

explanation: