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Question

2 oxides of a certain metal was separately heated in a current of hydrogen until constant weights were obtained. The water produced in each case was carefully collected and weighed. 2 grams of each oxide gave respectively 0.2517 grams and 0.4526 grams of water. Show that these results establish the law of multiple proportion.


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Solution

Step 1: Given data :

Weight of the each oxide = 2g
Weight of water produced in case 1 = 0.2517g
Weight of water produced in case 2 =0.4526g.

Step 2: Calculating the weight of oxygen in each oxide

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
18gof water is equivalent to 16gof oxygen
Thus 18g of water contains 16gof oxygen
Therefore, 0.2517gof water contains 16/18×0.2517=0.2237g
And, 0.4526g of water contains (16)/(18)×0.4526=0.4023g

Step 3: Calculating the weight of oxygen when combined with 1g of metal in each oxide

Now two cases arise :
Case 1
Weight of metal oxide = 2g
Weight of oxygen =0.2237g

Thus Weight of metal,

2-0.2237=1.7763g
Weight of oxygen combined with 1.7763g of metal = 0.2237g
and, Weight of oxygen combined with 1g of metal = (0.2237)/(1.7763)=0.1259g
Case 2
Now, Weight of metal oxide =2g
Weight of oxygen =0.4023g
Weight of metal = 2-0.4023=1.5977g
Weight of oxygen combined with 1g of metal =

(0.4023)/(1.5977)=0.2518g
Step 4: Comparing the weights of oxygen
The ratio of0.1259:0.2582 is the same as 1:2

This ratio explains the law of multiple proportions.


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