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Standard X

Chemistry

Balanced Chemical Equation

0.85 g of an ...

Question

$0.85 g$ of an organic compound is subjected to Kjeldahl's method for the estimation of nitrogen present. The ammonia evolved, neutralized $12.75 mL of 1 M H_{2} S O_{4}$ . The percentage of nitrogen in the compound is (to the nearest integer)

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42.00

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42.0

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Solution

$Milliequivalents of produced N H_{3} = Milliequivalents of consumed H_{2} S O_{4}$
$Percentage of nitrogen = \frac{14 \times 12.75 \times 2 \times 100}{1000 \times 0.85}$
$\Rightarrow Percentage of nitrogen = 42 %$

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0.85 g of an organic compound is subjected to Kjeldahl's method for the estimation of nitrogen present. The ammonia evolved, neutralized 12.75 mL of 1 M H2SO4. The percentage of nitrogen in the compound is (to the nearest integer)

Milliequivalents of produced NH3=Milliequivalents of consumed H2SO4 Percentage of nitrogen=14×12.75×2×1001000×0.85 ⇒Percentage of nitrogen=42%

$0.85 g$ of an organic compound is subjected to Kjeldahl's method for the estimation of nitrogen present. The ammonia evolved, neutralized $12.75 mL of 1 M H_{2} S O_{4}$ . The percentage of nitrogen in the compound is (to the nearest integer)

$Milliequivalents of produced N H_{3} = Milliequivalents of consumed H_{2} S O_{4}$
$Percentage of nitrogen = \frac{14 \times 12.75 \times 2 \times 100}{1000 \times 0.85}$
$\Rightarrow Percentage of nitrogen = 42 %$