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10 mL of hydrogen combine with 5mL of ox...

`10 mL` of hydrogen combine with `5mL` of oxygen to yield water. When 200 mL of hydrogen at N.T.P. are passed over heated `CuO`, the latter loses `0.144 g` of its mass. Do these results agree with the law of constant composition ?

Text Solution

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In first case : Mass of 10 mL of `H_(2)` at N.T.P. `= ((2g))/((22400mL))xx(10mL)=0.00089g`
Mass of 5mL of `O_(2)` at N.T.P. `= ((32g))/((22400mL))xx(5mL)=0.00714g`
Ratio of the masses of `H_(2) and O_(2) = 0.00089 : 0.00714 = 1:8`
In second case : Mass of 200 mL of `H_(2)` at N.T.P. `= ((2g))/((22400mL))xx(200mL)=0.0178g`
Loss in mass of CuO on heating is the same as that of oxygen `= 0.144 g`
`:.` Ratio by masses of `H_(2) and O_(2) = 0.0178 : 0.144 = 1:8`
Since the ratio in both the cases is the same, the results are in agreement with law of constant composition.
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