$0.85 g$ 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)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

42.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

42.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from

0.5

g of the compound in Kjeldahl's estimation of nitrogen neutralized

10

mL of

1 M H_{2} S O_{4.}

. The percentage of nitrogen in the compound is :

Q. During the estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from

0.5 g

of the compound in kjeldahls estimation of nitrogen, neutralized

10 m L

1 M

H_{2} {S O}_{4}

. Find out the percentage of nitrogen in the compound.

The ammonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M $H_{2} S O_{4}$ solution. The weight percentage of nitrogen in the compound is

Q. In the estimation of nitrogen by Kjeldahl's method,

2.8 g

for an organic compound required

20

millimoles of

H_{2} S O_{4}

for the complete neutralization of

N H_{3}

gas evolved. The percentage of nitrogen in the sample is:

Q. The amonia evolved from 2g of a compound in Kjeldahl's estimation of nitrogen neutralizes 10 mL of 2 M

H_{2} S O_{4}

solution. The weight percentage of nitrogen in the compound is

Join BYJU'S Learning Program

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)

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