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Krishna Subramanian
Asked: 2022-11-10T14:14:03+05:30 In:

One litre of mixture of `CO` and `CO_(2)` is passed through red hot charcoal in tube. The new volume becomes 1.4 litre. Find out % composition of mixture by volume. All measurements are made at same `P` and `T`

One litre of mixture of `CO` and `CO_(2)` is passed through red hot charcoal in tube. The new volume becomes 1.4 litre. Find out % composition of mixture by volume. All measurements are made at same `P` and `T`
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  1. Correct Answer – (a) `0.6`, (b) `24`
    After passing through red-hot charcoal, following reaction occurs
    `C(s)+CO_(2)(g) rarr 2CO(g)`
    If the `1.0 L` original mixture contain `x` litre of `CO_(2)`, after passing from the containing red-hot charcoal, the new volumes would be:
    `2x (`volume of `CO` obtained from `CO_(2))+1`
    `-x(“original “CO)=1+x=1.6` (given)
    `implies x=0.6`
    Hence original `1.0 L` mixture has `0.4 L CO` and `0.6L` of `CO_(2)` i.e. `40% CO` and `60% CO_(2)` by volume.
    (b) According to the given information, molecular formula of the compound is `M_(3)N_(2)`. Also, `1.0` mole of compound has `28 g` of nitrogen. If `X` is the molar mass of compound, then :
    `X xx 28/100=28`
    `implies X=100=3xx` Atomic weight of `M+28`
    `implies` Atomic weight of `M=72/3=24`

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