JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 31

Answer: (c) 15.72%

Explanation:
1. Pressure of dry nitrogen = 720 − 20 = 700 mm.
2. Volume at STP: \( V_0=\frac{700\times60\times273}{300\times760}=50.3\ \text{mL} \).
3. % N = \( \frac{28}{22400}\times\frac{50.3}{0.4}\times100=15.72\% \).

Teacher's Note:
1. Subtract the aqueous tension from the total pressure before converting to STP.
2. 22400 mL of N₂ at STP weighs 28 g.

Answer: (c) Liebig's method

Explanation:
1. The compound is burnt with excess CuO; H₂O is absorbed in anhydrous CaCl₂ and CO₂ in KOH solution.
2. This is Liebig's method. Kjeldahl's and Dumas' methods estimate nitrogen and the Carius method estimates halogens and sulphur.

Teacher's Note:
1. Match each method with the element it estimates.
2. Liebig's method uses weighed U-tubes of CaCl₂ and KOH.

Answer: (c) \( \ce{NH2NH2.HCl} \)

Explanation:
1. In the sodium fusion nitrogen is converted to NaCN only if the compound also contains carbon.
2. Hydrazine hydrochloride has no carbon, so no cyanide forms and the test fails. The other three compounds contain carbon.

Teacher's Note:
1. Think about what is needed to form NaCN during fusion.
2. A carbon-free nitrogen compound cannot give NaCN.

Answer: (a) \( \ce{Na4[Fe(CN)5NOS]} \)

Explanation:
1. \( \ce{Na2S + Na2[Fe(CN)5NO] -> Na4[Fe(CN)5NOS]} \) (violet).
2. So the coloured species is Na₄[Fe(CN)₅NOS].

Teacher's Note:
1. The sulphide ion adds to the nitroprusside complex.
2. Count the sodium ions needed to balance the charge: the complex ion is 4−.

Answer: (a) \( \ce{Na2S} \) and \( \ce{NaCN} \)

Explanation:
1. Sulphur gives Na₂S and nitrogen gives NaCN when fused with sodium.
2. NaSCN forms only when both N and S are present together.

Teacher's Note:
1. Sodium fusion converts covalent elements to ionic sodium salts.
2. Do not confuse the individual case with the case of both elements present.

Answer: (d) to decompose NaCN and \( \ce{Na2S} \) as HCN and \( \ce{H2S} \) so that they do not interfere with \( \ce{AgNO3} \)

Explanation:
1. Cyanide and sulphide would also precipitate with silver nitrate and mask the halide test.
2. Boiling with conc. HNO₃ removes them as volatile HCN and H₂S.

Teacher's Note:
1. Think about which ions react with Ag⁺ besides halides.
2. Volatile gases escape on boiling.

Answer: (a) \( \ce{N2} \), \( \ce{NH3} \)

Explanation:
1. Dumas' method: the compound is heated with CuO and the free nitrogen gas is measured.
2. Kjeldahl's method: the nitrogen is converted to (NH₄)₂SO₄ and then to NH₃, which is absorbed in standard acid.

Teacher's Note:
1. Recall the final form of nitrogen in each method.
2. Dumas measures a gas volume; Kjeldahl uses titration.

Answer: (c) sulphur

Explanation:
1. \( \ce{Na2S + (CH3COO)2Pb -> PbS + 2CH3COONa} \)
2. PbS is black, which shows sulphur.

Teacher's Note:
1. Black lead sulphide is the clue.
2. The acetic acid prevents precipitation of other lead salts.

Answer: (d) a blood-red colour

Explanation:
1. Both elements form sodium thiocyanate, NaSCN, during fusion.
2. With FeCl₃ it gives \( \ce{Fe(SCN)3} \), which is blood red, instead of Prussian blue.

Teacher's Note:
1. Thiocyanate replaces cyanide when both N and S are present.
2. Fe(SCN)₃ is the classic blood-red complex.

Answer: (a) X = NaCN, Y = \( \ce{Na2S} \)

Explanation:
1. Sodium is a strong reducing agent, so sulphur is reduced to sulphide (Na₂S), not oxidised to sulphate.
2. Nitrogen (with carbon) forms sodium cyanide, NaCN.

Teacher's Note:
1. Fusion with sodium reduces the elements.
2. Cyanide, not isocyanide, is formed.

Answer: (d) convert nitrogen, sulphur or halogen, if present, into soluble ionic sodium compounds

Explanation:
1. Covalently bonded N, S and halogens cannot be tested directly.
2. Fusion with sodium turns them into ionic sodium salts (NaCN, Na₂S, NaX) which dissolve in water and can be identified.

Teacher's Note:
1. The test needs the elements as ions in solution.
2. Water extraction of the fused mass gives the sodium extract.

Answer: (a) 28

Explanation:
1. Acid taken: 50 mL of 0.5 M H₂SO₄ = 50 mL of 1 N = 50 milliequivalents.
2. Unused acid = 80 × 0.5 = 40 meq, so acid used by NH₃ = 10 meq.
3. Mass of N = \( \frac{14\times10}{1000}=0.14\ \text{g} \); % N = \( \frac{0.14}{0.5}\times100=28 \).

Teacher's Note:
1. Convert molarity of H₂SO₄ to normality (×2).
2. Used acid in meq equals meq of NH₃ and hence of nitrogen.

Answer: (b) 46.67

Explanation:
1. Milliequivalents of acid used = 20 × 1 = 20.
2. Mass of N = \( \frac{14\times20}{1000}=0.28\ \text{g} \).
3. % N = \( \frac{0.28}{0.6}\times100=46.67 \).

Teacher's Note:
1. 1 meq of NH₃ contains 14 mg of nitrogen.
2. Divide by the sample mass.

Answer: (c) 20%

Explanation:
1. Molar mass of BaSO₄ = 137 + 32 + 64 = 233.
2. Mass of S = \( \frac{32}{233}\times0.699=0.096\ \text{g} \).
3. % S = \( \frac{0.096}{0.480}\times100=20 \).

Teacher's Note:
1. Each 233 g of BaSO₄ contains 32 g of sulphur.
2. Divide the sulphur mass by the sample mass.

Answer: (d) 10

Explanation:
1. Acid taken = 60 mL of 0.2 N = 12 meq (M/10 H₂SO₄ is 0.2 N). NaOH used = 20 × 0.1 = 2 meq.
2. Acid used by NH₃ = 12 − 2 = 10 meq, so mass of N = \( \frac{14\times10}{1000}=0.14\ \text{g} \).
3. % N = \( \frac{0.14}{1.4}\times100=10 \).

Teacher's Note:
1. M/10 H₂SO₄ is 0.2 N because it is dibasic.
2. Subtract the unused milliequivalents.

Answer: (b) 16.45

Explanation:
1. Pressure of dry N₂ = 715 − 15 = 700 mm.
2. Volume at STP: \( \frac{700\times55\times273}{300\times760}=46.1\ \text{mL} \).
3. % N = \( \frac{28}{22400}\times\frac{46.1}{0.35}\times100\approx16.45 \).

Teacher's Note:
1. Correct the pressure for aqueous tension first.
2. Then reduce the volume to STP and apply the nitrogen formula.