A 0 . 166   g sample of an organic compound was digested with conc. H 2 SO 4 and then distilled with NaOH . The ammonia gas evolved was passed through 50 . 0   mL of 0 . 5   N   H 2 SO 4 . The used acid required 30 . 0   mL of 0 . 25   N   NaOH for complete…

Question

A 0 . 166 &#160; g sample of an organic compound was digested with conc. H 2 SO 4 and then distilled with NaOH . The ammonia gas evolved was passed through 50 . 0 &#160; mL of 0 . 5 &#160; N &#160; H 2 SO 4 . The used acid required 30 . 0 &#160; mL of 0 . 25 &#160; N &#160; NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is

Aicharya · Accepted Answer

Complete acid is not used in the reaction.

Milli eq. of ${NH}_{3} =$ milli eq. of $NaOH$

$= 0.25 \times 30$

$= 7.5$

Millimole of Nitrogen $= 7.5$

$W_{Nitrogen} = 7.5 \times 14 \times 10^{- 3}$

$= 0.105 gram$

$%$ of Nitrogen $= \frac{0.105}{0.166} \times 100 = 63.25 = 63$

A 0 . 166   g sample of an organic compound was digested with conc. H 2 SO 4 and then distilled with…

Solution

A 0 . 166 &#160; g sample of an organic compound was digested with conc. H 2 SO 4 and then distilled with…

Solution

A 0 . 166 g sample of an organic compound was digested with conc. H 2 SO 4 and then distilled with…