0.44 gm of organic compound `C_(x) H_(y)` O which occupied 224 ml at N – InterviewSolution

InterviewSolution

1.	0.44 gm of organic compound `C_(x) H_(y)` O which occupied 224 ml at NTP and on combustion gave 0.88 gm `CO_(2)`. The ratio of X to Y in the compound 1sA. `1:1`B. `1:2`C. `1:3`D. `1:4`
Answer» Correct Answer - B 224 mL at NTP of O.C weighs = `044` g `therefore` 22400 ml at NTP of O.C. weighs = `(044)/(224) xx 22400 = 44 g ` Wt. of carbon 1 mol (44 g) of the O.C. = `(12)/(44) xx 088 xx (44)/(044) = 24 `g = 2g atom of carbon . `therefore` X = 2 As the molecular mass of the compound is 44 . Therefore molecular formula can be `C_(2)H_(4)O` Hence Y = 4 , `therefore X : Y = 2 : 4 = 1 : 2 `

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