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6g of the organic compound on heating with $N a O H$ gives $N H_{3}$ which is neutralized by 200ml of $1 N H C l$ percentage of nitrogen is:

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46.67%

12%

26.67%

60%

answer is C.

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Detailed Solution

$% N = \frac{1.4 \times 200}{6} = 46.67 %$

$% N = \frac{1.4 \times N V}{w t . o f o r g . c o m}$

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