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InterviewSolution
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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 a metal by weight. Three atoms of the metal combine with two atoms of N. find the atomic weight of the metal. |
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Answer» (a). Let the volume of CO in the mixture be V. The volume of `CO_(2)` in the mixture is `(1-V)` `CO_(2)+Cto2CO` `(1-V)LCO_(2)` forms `2(1-V)LCO`. Total CO, `V+2(1-V)=1.6` or `V=0.4` `CO=0.4L` `CO_(2)=1-V=0.6L` (b). 28 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. " Eq of "`N=(28)/((14)/(3))=(1)/(6)` " Eq of "metal `=(72)/((M)/(2))=(144)/(M)` `therefore(1)/(6)=(144)/(M)impliesM=(144)/(6)=24g` |
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