0.90 gm. of a solid organic compound(molecular weight = 90) containing

1.	0.90 gm. of a solid organic compound(molecular weight = 90) containing C, H and Owas heated with oxygen having volume 224 ml.at NTP. After the reaction, total volume of gas-ses was 560 ml. at NTP. On treatment withKOH, the volume decreases to 112 ml.Determined the molecular formula.
Answer» Explanation: Number of moles of organic compound is = 0.9/90 = 0.01 mol After KOH treatment volume CO2 decreased = 560-112 = 448 ML It MEANS the volume of CO2 produced is 448 ml 22400 ml of CO2 = 1 Mol of CO2 at stp So 448 ml contains = 448/22400 = 0.02 mol 0.01 mol of organic compound produces 0.02 mol CO2 So 1 mol organic compound produce = 0.02/0.01 = 2 mol CO2 So number of C in the organic compound is 2 So the molecular formula C2HxOy 24+ x1 + 16 y = 90 Substitute some POSSIBLE values gives x =2 and y = 4 Hence the molecular formula is C2H2O4

0.90 gm. of a solid organic compound(molecular weight = 90) containing C, H and Owas heated with oxygen having volume 224 ml.at NTP. After the reaction, total volume of gas-ses was 560 ml. at NTP. On treatment withKOH, the volume decreases to 112 ml.Determined the molecular formula.

Answer»

Explanation:

Number of moles of organic compound is = 0.9/90 = 0.01 mol

After KOH treatment volume CO2 decreased = 560-112 = 448 ML

It MEANS the volume of CO2 produced is 448 ml

22400 ml of CO2 = 1 Mol of CO2 at stp

So 448 ml contains = 448/22400 = 0.02 mol

0.01 mol of organic compound produces 0.02 mol CO2

So 1 mol organic compound produce = 0.02/0.01 = 2 mol CO2

So number of C in the organic compound is 2

So the molecular formula C2HxOy 24+ x1 + 16 y = 90

Substitute some POSSIBLE values gives x =2 and y = 4

Hence the molecular formula is C2H2O4

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