Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M . The percentage of nitrogen in the compound is ______%.

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]

Ans.
(56)

   $Millimoles of sulphuric acid (n_{H_{2} {SO}_{4}}) = Molarity \times volume in mL$

   $= 2 M \times 10 mL = 20 mmol$

   $Ammonia is a base, it reacts with sulphuric acid to form salt:$

   $2 {NH}_{3} + H_{2} {SO}_{4} \to {({NH}_{4})}_{2} {SO}_{4}$

   $By stoichiometry:$

   $\frac{n_{N H_{3}}}{2} = \frac{n_{H_{2} S O_{4}}}{1}$

   $\frac{n_{N H_{3}}}{2} = \frac{20 mmol}{1}$

   $n_{N H_{3}} = 40 mmol = 0.04 mol$

   $Moles of N (n_{N}) = Moles of ammonia (n_{N H_{3}}) = 0.04 mol$

   $Since amount of nitrogen is conserved, number of moles of N in organic compound (n_{N}) = number of moles of N in ammonia$

   $(n_{N H_{3}}) = 0.04 mol$

   $Weight of N in organic compound (W_{N}) = n_{N} \times Molar mass of N$

   $= 0.04 mol \times 14 {g mol}^{- 1} = 0.56 g$

   $Percentage of nitrogen in organic compound:$

   $= \frac{W_{N}}{W_{compound}} \times 100 = \frac{0.56}{1} \times 100 = 56 %$

Want Us to Email you About Special Offers & Updates?