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States Of Matter

Question
CBSEENCH11008308

In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be

(Aqueous tension at 300 k = 15 mm)

  • 16.45

  • 17.45

  • 14.45

  • 15.45

Solution

A.

16.45

Gas equation,
fraction numerator straight p subscript 1 straight V subscript 1 over denominator straight T subscript 1 space end fraction space equals space fraction numerator straight p subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction
Where comma space straight p subscript 2 space equals space pressure space of space straight N subscript 2 space at space STP space equals space 760 space mm
straight T subscript 2 space equals space Temperature space of space straight N subscript 2 space straight t space STP space equals space 273 straight K
straight V subscript 2 space equals space ?
Volume space of space straight N subscript 2 space at space STP space left parenthesis by space Gas space equation right parenthesis
open parentheses fraction numerator straight rho space minus space straight rho subscript 1 over denominator straight t plus 273 end fraction close parentheses straight V subscript 1 space straight x space 273 over 160 space equals space straight V subscript 2
Where space straight p subscript 1 space equals space straight rho space minus space straight rho subscript 1
straight rho space equals space 715 space mm space left parenthesis pressure space at space which space straight N subscript 2 space collected right parenthesis
straight rho subscript 1 space equals space aqueous space tension space of space water space equals 15 mm
straight T subscript 1 space equals space straight t space plus space 273 space equals space 300 space straight k space
straight V subscript 1 space equals space 55 space mL space space equals space volume space of space moist space nitrogen space in space nitrometer
therefore space straight V subscript 2 space equals space fraction numerator left parenthesis 715 minus 15 right parenthesis space straight x space 55 over denominator 300 end fraction space straight x space 273 over 760 space equals space 40.098 space mL
percent sign space of space nitrogen space in space given space compound

equals space 28 over 22400 space straight x space straight V subscript 2 over straight W space straight x space 100 space

equals space 2 over 22400 space straight x space fraction numerator 46.09 over denominator 0.35 end fraction space straight x space 100 thin space

space equals space 16.45 percent sign