In Kjeldahl's method, ammonia from 5 g of food neutralizes 30 cm3 of 0.1 N acid. The percentage of nitrogen in the food is

Question

In Kjeldahl's method, ammonia from 5 g of food neutralizes 30 cm3 of 0.1 N acid. The percentage of nitrogen in the food is (A) 8.4 (B) 0.84 (C) 1.68 (D) 16.8

Tardigrade Admin · Accepted Answer

From Kjeldahl's method, percentage of nitrogen =(1.4 × N × V/w)=(1.4 × 0.1 × 30/5) =0.84 %

Q. In Kjeldahl's method, ammonia from $5 g$ of food neutralizes $30 c m^{3}$ of $0.1 N$ acid. The percentage of nitrogen in the food is

8.4

0.84

1.68

16.8

From Kjeldahl's method,

percentage of nitrogen

$= \frac{1.4 \times N \times V}{w} = \frac{1.4 \times 0.1 \times 30}{5}$

$= 0.84%$

Q. In Kjeldahl's method, ammonia from 5g of food neutralizes 30cm3 of 0.1N acid. The percentage of nitrogen in the food is

8.4

0.84

1.68

16.8

From Kjeldahl's method, percentage of nitrogen =w1.4×N×V​=51.4×0.1×30​ =0.84%

Q. In Kjeldahl's method, ammonia from $5 g$ of food neutralizes $30 c m^{3}$ of $0.1 N$ acid. The percentage of nitrogen in the food is

From Kjeldahl's method,

percentage of nitrogen

$= \frac{1.4 \times N \times V}{w} = \frac{1.4 \times 0.1 \times 30}{5}$

$= 0.84%$