Question

In Duma's method for estimation of nitrogen 0.25 g of an organic compound gave 40 ml of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm the percentage of nitrogen in the compounds is,

17.36

18.20

16.76

15.76

Solution

16.76

straight p subscript straight N subscript 2 space end subscript space equals straight P subscript total space minus straight p subscript straight H subscript 2 straight O end subscript space equals 725 minus 25 space equals 700 space mm According space to space Duma apostrophe straight s space method space for space estimation space of space straight N comma weight space of space nitrogen space equals pVM over RT where comma space straight p space is space pressure space equals space 700 mm space straight V space of space nitrogen equals 40 space mL straight M space molecular space mass space of space straight N subscript 2 space equals space 28 straight R minus 0.0821 straight T minus 300 straight K space equals space fraction numerator 700 space straight x 40 space straight x 28 over denominator 760 space straight x space 1000 space straight x 300 space straight x space 0.0821 end fraction space equals space 0.042 space straight g therefore comma space percent sign space of space nitrogen space space equals space fraction numerator 0.042 over denominator 0.25 end fraction space straight x space 100 space equals space 16.573 percent sign

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