You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Types of Redox Reactions

In Kjeldahl's...

Question

In Kjeldahl's method, ammonia from 5 g of good neutralizes 30 $c m^{2}$ of 0.1 N acid. The percentage of nitrogen in the food is:

0.84

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

8.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.68

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 0.84
From Kjeldahl's meghod, percentage of nitrogen
$= \frac{1.0 \times N \times V}{w} = \frac{1.4 \times 0.1 \times 30}{5}$
$= 0.84 %$

Suggest Corrections

Similar questions

Q. In Kjeldahl's method, ammonia from

5

g of substrance neutralise

30 c m^{3}

0.1 N

acid. The percentage of nitrogen in the substrance is:

Q. 0.303 g of an organic compound was analysed for nitrogen by Kjeldahl’s method. The ammonia gas evolved was absorbed in 50 ml of 0.1N-H2SO4. The excess acid required 25 ml of 0.1 N NaOH for neutralization. The percentage of nitrogen in the given compound is

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 :

29.5

mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in

20

ml of

0.1

M

HCl solution. The excess of the acid required

15

ml of

0.1

M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is:

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

Join BYJU'S Learning Program

In Kjeldahl's method, ammonia from 5 g of good neutralizes 30 cm2 of 0.1 N acid. The percentage of nitrogen in the food is:

The correct option is D 0.84From Kjeldahl's meghod, percentage of nitrogen=1.0×N×Vw=1.4×0.1×305=0.84%

In Kjeldahl's method, ammonia from 5 g of good neutralizes 30 $c m^{2}$ of 0.1 N acid. The percentage of nitrogen in the food is:

The correct option is D 0.84
From Kjeldahl's meghod, percentage of nitrogen
$= \frac{1.0 \times N \times V}{w} = \frac{1.4 \times 0.1 \times 30}{5}$
$= 0.84 %$