The percentage of nitrogen in urea $(N H_{2} C O N H_{2})$ , is:

38.4

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46.6

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59.1

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61.3

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Solution

The correct option is C $46.6$
ANS; B
molecular mass of urea = 60
molecular mass of urea = 28
% of nitrogen in urea = $\frac{28 \times 100}{60}$ = 46.6

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