Chemistry · Practical Organic Chemistry

JEE Main 2024 — 1 February, Shift 2 — Question 80

Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2MH2SO42 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}. The percentage of nitrogen in the compound is \qquad %\%.

Answer: 56

Numerical answer — enter this value.

Step-by-step solution

H2SO4+2NH3→(NH4)2SO4\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NH}_{3} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}

Millimole of H2SO4→10×2\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 10 \times 2

So Millimole of NH3=20×2=40\mathrm{NH}_{3}=20 \times 2=40

∴\therefore Mole of N=401000\mathrm{N}=\frac{40}{1000}

wt. of N=401000×14\mathrm{N}=\frac{40}{1000} \times 14 %\%

composition of N in organic compound =40×141000×1×100=\frac{40 \times 14}{1000 \times 1} \times 100 =56%=56 \%

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
Practical Organic Chemistry
Topic
Quantitative organic analysis
Following Kjeldahl's method, 1 g of organic compound released… | JEE Main 2024 PYQ with Solution · DhiX AI