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1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098 g In another experiment, 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g. Show that these result illustrate the law of constant composition.

Answer» <html><body><p></p>Solution : In the first <a href="https://interviewquestions.tuteehub.com/tag/experiment-980358" style="font-weight:bold;" target="_blank" title="Click to know more about EXPERIMENT">EXPERIMENT</a>: <br/> Mass of CuO taken = 1.375 g <br/> Mass of Cu obtained = 1.098 g <br/> `therefore` Mass of oxygen that combined with <br/> `Cu = 1.375 - 1.098 = 0.277g`<br/> Thus, the percentage of oxygen in the given sample of<br/> `CuO = 0.277/(1.375)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> = 20.14 %` <br/> In the serial experiment:<br/> Since, 1.178 g of Cu yielded 1.467 g of CuO, the mass of oxygen added would be: <br/> `=1.476 - 1.178 = 0.298 g` <br/> `therefore` Percentage of oxygen in this <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> <br/> `=(0.298 xx 100)/1.476 = 20.19%` <br/> Since, the percentage of oxygen is almost the same (within the limit of experimental errors) in the two cases, the data are in accordance to the <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of constant proportion and prove it.</body></html>


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