Two oxides of certian metal were separetly heated in a current of hydrogen untill constant weights were obtained . The.water produces 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 establosh the law of multiple proportions.

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Qn:2 oxidesof a certan metal wer separately heated in a current of h2 until constant wt.s were obtained. d h2o produced in each case was carefully collected & weighted. it was observed dat 1g of each oxide gave 0.1254g &0.2263g of h2o resp.show it illustrates law of multiple proportn
An:
(1) Calculation of oxygen in each oxide

18 g of water = 16 g of O2

0.1254 g water in first oxide = 16/180.1254 = 0.111g oxygen

similarly 0.226 g water in second oxide = 16/180.2263 = 0.20 g oxygen

(2) Calculation of weight of oxygen which combine with 1 g of metal in each oxide

weight of metal oxide = 1.0 g ; weight of oxygen = 0.111g

Weight of metal = 1.0 – 0.111 = 0.889 g

Since 0.889 g of metal combine with 0.111 g of oxygen, therefore

1.0 g metal will combine with (0.111/0.889)* 1.0 = 0.124 g of oxygen

Similarly Weight of metal in second oxide = 1.0 – 0.2 = 0.8 g

Since 0.8 g of metal combine with 0.2 g of oxygen, therefore

1.0 g metal will combine with (0.2/0.8)* 1.0 = 0.25 g of oxygen

3) Ratio of oxygen for the same mass (1g) of metal

0.124 g : 0.25 g

or 1 : 2

This is a simple whole number ratio. Thus law of multiple proportions is verified.

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