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 = $2\mathrm{g}$

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

$18g$of water is equivalent to $16g$of oxygen

Thus $18g$ of water contains $16g$of oxygen

Therefore, $0.2517g$of water contains $16/18\times 0.2517=0.2237g$

And, $0.4526g$ of water contains $\left(16\right)/\left(18\right)\times 0.4526=0.4023g$

**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,

$2-0.2237\phantom{\rule{0ex}{0ex}}=1.7763g$

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 of$0.1259:0.2582$ is the same as $1:2$

This ratio explains the law of multiple proportions.

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