Question

An element forms two oxides. Two gram of each
oxide on heating
with H₂ gives 0.2517 g
and
0.4526 g water. show that these data confirms the law
of multiple proportion.
​

Answer 1

The law of multiple proportion in all the given cases can be verified by combining the weight of oxygen with the weight of the metal in two different oxides. Therefore, following are the steps that needs to be followed:

Step 1: To calculate the weight of oxygen in each oxide

Weight of the each oxide = 2g

Weight of water produced in case 1 = 0.2517g

Weight of water produced in case 2 = 0.4526g

18g of H2O is equivalent to 16g of oxygen

Which means that 18g of water contains 16g of oxygen

Therefore, 0.2517g of water contains (16)/(18)*0.2517 = 0.2237g

And, 0.4526g of water contains (16)/(18)*0.4526 = 0.4023g

Step 2: To calculate the weight of oxygen when combined with 1g of metal in each oxide

Case 1

Weight of metal oxide = 2g

Weight of oxygen =0.2237g

Weight of metal = 2-0.2237 =1.7763g

Weight of oxygen combining with 1.7763g of metal = 0.2237g

Weight of oxygen combining with 1g of metal = (0.2237)/(1.7763) = 0.1259g

Case 2

Weight of metal oxide = 2g

Weight of oxygen =0.4023g

Weight of metal = 2-0.4023 =1.5977g

Weight of oxygen combining with 1g of metal = (0.4023)/(1.5977) = 0.2518g

Case 3:

To compare the weights of oxygen

The ratio of 0.1259:0.2582 is same as 1:2

This ratio explains the law of multiple proportions.

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Answer 2

Explanation:

To verify the law of multiple proportion in each case, the weight of oxygen combining with the fixed weight of the metal in the two different oxides should bear a simple ratio to one another. So, to prove this, we move step wise as follows:

Step 1. Calculate the weight of oxygen in each oxide.

Here, we are given:

Weight of each oxide = 2.0 g

Weight of water produced in case I = 0.2517 g

Weight of water produced in case II = 0.4526 g

18 g of H2O ≡ 16 g of oxygen

i.e.18 g of water contain oxygen = 16 g

∴ 0.2517 g of water contains oxygen

& 0.4526 g of water contains oxygen

Step 2. Calculate the weight of oxygen which would combine with 1 g of metal in each oxide.

In case I, Weight of metal oxide = 2g and weight of oxygen = 0.2237 g

∴ Weight of metal = 2 – 0.2237 = 1.7763 g

∴ Weight of oxygen which combines with 1.7763 g of metal = 0.2237 g

∴ Weight of oxygen which combines with 1 g of metal = 0.2237/1.7763 g = 0.1259 g

In case II, Weight of metal oxide = 2g and weight of oxygen = 0.4023 g

∴ Weight of metal = 2 – 0.4023 = 1.5977 g

Weight of oxygen which combines with 1.5977 g of metal = 0.4023 g

∴ Weight of oxygen which combines with 1 g of metal = 0.4023/1.5977 g = 0.2518 g

Step 3. Compare the weights of oxygen which combine with the same weight of metal in the two oxides.

The weights of oxygen which combine with 1 g of metal in the two oxides are respectively 0.1259 g and 0.2518 g. These weights are in the ratio 0.1259 : 0.2518 or 1 : 2.

Since this is a simple ratio, so the above results establish the law of multiple proportions.

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