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(a). Take 1 L of a mixture of CO and CO_(2) Pass this mixture through a tube containing red hot charcoal. The volume now becomes 1.6 L. The volumes are measured under the same conditions. Find the composition of the mixture by volume. (b). A compound contains 28 percent of nitrogen 72 percent of ametal by weight. Three atoms of the metal combine with two atoms of N. find the atomic weight of the metal.

Answer» <html><body><p></p>Solution :(a). Let the volume of CO in the mixture be <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>. <br/> The volume of `CO_(2)` in the mixture is `(1-V)` <br/> `CO_(2)+Cto2CO` <br/> `(1-V)LCO_(2)` forms `2(1-V)LCO`. <br/> Total CO, <br/> `V+2(1-V)=1.6` <br/> or `V=0.4` <br/> `CO=0.4L` <br/> `CO_(2)=1-V=0.6L` <br/> (b). <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a> g of nitrogen combines with 72 g of metal. three atoms of metal. Three atoms of metal combine with two atoms of nitrogen. This means the valency of metal is 2. <br/> " Eq of "`<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=(28)/((14)/(3))=(1)/(6)` <br/> " Eq of "metal `=(72)/((M)/(2))=(144)/(M)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>(1)/(6)=(144)/(M)impliesM=(144)/(6)=24g`</body></html>


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