1.0 g each of certain metal oxide on separate heating in a current of hydrogen produced 0.126g and 0.226g of water . Show that these results vsrify the laws of multiple propertion .
(I)Calculation of oxygen in each oxide
(ii)Calculation of weight of oxygen which combine with 1 g of metal in each oxide.
Answers
(i)Calculation of oxygen in each oxide
18g of water=16g oxygen
0.126 water in first oxide = 16/18×0.126=0.112g oxygen
similarly,0.226g water in second oxide= 16/18× 0.226=0.20g oxygen
(ii)Calculation of weight of oxygen which combine with 1 g of metal in each oxide
Wt. of metal oxide =1.0g Weight of oxygen=0.112g
weight of metal =1.0-0.112g=0.888g
since 0.888g metal combine with 0.112g oxygen
1.0g metal will combine with 0.112/0.888×1.0=0.126 g oxygen
similarly weight of metal in second oxide =1.0-0.2=0.8 g
0.8 metal combine with 0.2 g oxygen
1.0 g metal will combine with 0.2/0.8×1.0=0.25 g
Answer:
(i)Calculation of oxygen in each oxide
18g of water=16g oxygen
0.126 water in first oxide = 16/18×0.126=0.112g oxygen
similarly,0.226g water in second oxide= 16/18× 0.226=0.20g oxygen
(ii)Calculation of weight of oxygen which combine with 1g of metalin each oxide
Wt. of metal oxide =1.0g Weight of oxygen=0.112g
weight of metal =1.0-0.112g=0.888g
since 0.888g metal combine with 0.112g oxygen
1.0g metal will combine with 0.112/0.888×1.0=0.126 g oxygen
similarly weight of metal in second oxide =1.0-0.2=0.8 g
0.8 metal combine with 0.2 g oxygen
1.0 g metal will combine with 0.2/0.8×1.0=0.25 g