Practical Organic Chemistry — JEE Main practice

33 questions

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Sample questions with solutions

Q1 · 2026

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g mol $^{-1}$ ). Molar mass of barium sulphate is 233 g mol $^{-1}$ .

A.
4.55%
B.
16.48%
C.
10.30%
D.
21.97%

Answer: D

Calculate the moles of BaSO $_4$ formed. Given

n(\mathrm{BaSO}_4) = \frac{1.2}{233}\ \mathrm{mol}

Determine the moles (and mass) of sulphur. Since there's 1 S atom per formula unit,

n(\mathrm{S}) = \frac{1.2}{233}\ \mathrm{mol} \quad \Rightarrow \quad m(\mathrm{S}) = \frac{1.2}{233}\times 32 = 0.1648\ \mathrm{g}

Calculate the percentage of sulphur in the 0.75 g compound sample:

\%\mathrm{S} = \frac{0.1648}{0.75}\times 100 = 21.97\%

Hence, the answer is option D.

Q2 · 2026

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer)

(Given, molar mass in g mol $^{-1}$ : O = 16, Mg = 24, P = 31)

A.
40
B.
30
C.
50
D.
20

Answer: C

Determine the molar mass of magnesium pyrophosphate ( $\mathrm{Mg}_2\mathrm{P}_2\mathrm{O}_7$ ). Given

M = 2(24)+2(31)+7(16) = 48+62+112 = 222\ \mathrm{g/mol}

Determine the mass of phosphorus per mole. Since there are 2 P atoms per formula unit,

m(\mathrm{P}) = 2\times 31 = 62\ \mathrm{g}\ \text{per } 222\ \mathrm{g\ Mg}_2\mathrm{P}_2\mathrm{O}_7

Scale to the actual mass obtained (1.79 g). Given

m(\mathrm{P}) = \frac{62}{222}\times 1.79 \approx 0.4999\ \mathrm{g}

Calculate the percentage of phosphorus in the 1.00 g compound sample:

\%\mathrm{P} = \frac{0.4999}{1.00}\times 100 \approx 49.99\% \approx 50\%

Hence, the answer is option C.

Q3 · 2026

When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M $\mathrm{H}_2\mathrm{SO}_4$ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is:

A.
42
B.
0.21
C.
21
D.
0.42

Answer: A

Calculate the equivalents of $\mathrm{H}_2\mathrm{SO}_4$ used. Since $\mathrm{H}_2\mathrm{SO}_4$ is dibasic (2 equivalents per mole),

\text{equivalents} = \frac{15\times 1\times 2}{1000}

Relate this to moles of ammonia (and thus nitrogen). Given ammonia is monobasic, moles of $\mathrm{NH}_3$ = equivalents of acid neutralized,

\text{moles of N} = \frac{15\times 1\times 2}{1000} = 0.03\ \mathrm{mol}

Calculate the mass of nitrogen. Given

\text{mass of N} = 0.03\times 14 = 0.42\ \mathrm{g}

Calculate the percentage of nitrogen in 1 g of compound:

\%\mathrm{N} = \frac{0.42}{1}\times 100 = 42\%

Hence, the answer is option A.

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