In the Duma’s method for the estimation of nitrogen, 0.84 g of an organic compound gave 448 mL of nitrogen at STP. The % of N in the compound is:

33.3 %

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66.7 %

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50 %

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60 %

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Solution

The correct option is B $66.7 %$
$% o f N = \frac{28 \times V \times 100}{22400 \times W} = \frac{28 \times 448 \times 100}{22400 \times 0.84} = 66.66 % = 66.7 %$

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Q. In the Dumas method for the estimation of nitrogen, 0.84 g of an organic compound Gave 448 mL of nitrogen at STP. The percentage of nitrogen in the compound is:

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Q. In the Dumas method for the estimation of nitrogen, 0.84 g of an organic compound Gave 448 mL of nitrogen at STP. The percentage of nitrogen in the compound is:

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In the Duma’s method for the estimation of nitrogen, 0.84 g of an organic compound gave 448 mL of nitrogen at STP. The % of N in the compound is:

The correct option is B 66.7%% of N=28×V×10022400×W=28×448×10022400×0.84=66.66%=66.7%

The correct option is B $66.7 %$
$% o f N = \frac{28 \times V \times 100}{22400 \times W} = \frac{28 \times 448 \times 100}{22400 \times 0.84} = 66.66 % = 66.7 %$