JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 11

Answer: (b) 28%

Explanation:
1. 100 mL of N/10 acid equals 10 mL of 1 N acid. 30 mL of N/10 NaOH equals 3 mL of 1 N alkali.
2. Acid consumed by ammonia = 10 - 3 = 7 mL of 1 N.
3. \( \%N = \dfrac{1.4\times1\times7}{0.35} = 28 \).

Teacher's Note:
1. Subtract the unused acid from the total acid taken, in the same normality.

Answer: (a) 17.5%

Explanation:
1. Pressure of dry nitrogen = 715 - 15 = 700 mm.
2. Volume at STP \( = \dfrac{273\times700\times50}{300\times760} = 41.9 \) mL.
3. Mass of N2 \( = \dfrac{28\times41.9}{22400} = 0.0524 \) g; \( \% = \dfrac{0.0524}{0.3}\times100 = 17.46 \).

Teacher's Note:
1. Do not forget to subtract the aqueous tension.

Answer: (d) 56%

Explanation:
1. 10 mL of 1 M \( H_2SO_4 \) = 20 mL of 1 N acid = 20 mL of 1 M \( NH_3 \).
2. Nitrogen \( = \dfrac{14\times20}{1000} = 0.28 \) g; \( \% = 0.28/0.5\times100 = 56 \).

Teacher's Note:
1. H2SO4 is dibasic, so 1 M means 2 N.

Answer: (c) 34.0%

Explanation:
1. Molar mass of AgBr = 108 + 80 = 188 g.
2. Bromine in 0.12 g AgBr \( = \dfrac{80\times0.12}{188} = 0.0511 \) g.
3. \( \% = \dfrac{0.0511}{0.15}\times100 = 34.04 \).

Teacher's Note:
1. Use the atomic mass of the halogen over the molar mass of the silver halide.

Answer: (b) 42.1%

Explanation:
1. Molar mass of \( BaSO_4 = 137 + 32 + 64 = 233 \) g.
2. Sulphur \( = \dfrac{32\times0.4813}{233} = 0.0661 \) g.
3. \( \% = \dfrac{0.0661}{0.157}\times100 = 42.1 \).

Teacher's Note:
1. One mole of BaSO4 contains one mole (32 g) of sulphur.

Answer: (a) 60% and 10%

Explanation:
1. \( \%C = \dfrac{12}{44}\times\dfrac{0.44}{0.2}\times100 = 60 \).
2. \( \%H = \dfrac{2}{18}\times\dfrac{0.18}{0.2}\times100 = 10 \).

Teacher's Note:
1. 12/44 of the CO2 mass is carbon and 2/18 of the water mass is hydrogen.

Answer: (c) 53.3%

Explanation:
1. Carbon \( = \dfrac{12}{44}\times0.44 = 0.12 \) g; hydrogen \( = \dfrac{2}{18}\times0.18 = 0.02 \) g.
2. Oxygen = 0.30 - 0.12 - 0.02 = 0.16 g.
3. \( \%O = \dfrac{0.16}{0.30}\times100 = 53.3 \).

Teacher's Note:
1. Oxygen is found by difference when no direct method is used.

Answer: (b) ferric ferrocyanide

Explanation:
1. \( 6NaCN + FeSO_4 \to Na_4[Fe(CN)_6] \), and \( 3Na_4[Fe(CN)_6] + 4FeCl_3 \to Fe_4[Fe(CN)_6]_3 \) (Prussian blue).

Teacher's Note:
1. Ferric ferrocyanide, Fe4[Fe(CN)6]3, is the blue compound.

Answer: (a) a blood red colour of \( Fe(SCN)_3 \)

Explanation:
1. Sodium thiocyanate forms: \( Na + C + N + S \to NaSCN \).
2. \( 3NaSCN + FeCl_3 \to Fe(SCN)_3 + 3NaCl \): blood red.

Teacher's Note:
1. The thiocyanate forms when the fusion is done without excess sodium.

Answer: (d) sulphur

Explanation:
1. \( Na_2S + Na_2[Fe(CN)_5NO] \to Na_4[Fe(CN)_5NOS] \) (violet).

Teacher's Note:
1. Lead acetate test also detects sulphur, giving black PbS.

Answer: (c) remove \( Na_2S \) and NaCN as \( H_2S \) and HCN

Explanation:
1. Sulphide and cyanide would otherwise precipitate with \( Ag^+ \) as \( Ag_2S \) and AgCN and interfere.

Teacher's Note:
1. Boil in a fume cupboard because H2S and HCN are toxic.

Answer: (b) bromine

Explanation:
1. AgCl is white and soluble in ammonia; AgBr is pale yellow and sparingly soluble; AgI is yellow and insoluble.

Teacher's Note:
1. AgF is soluble in water.

Answer: (a) phosphorus

Explanation:
1. Phosphorus is converted to phosphate with Na2O2 and gives ammonium phosphomolybdate, a canary yellow precipitate.

Teacher's Note:
1. Arsenic gives a similar test, but this is the standard test for phosphorus.

Answer: (d) turns blue on absorbing water

Explanation:
1. \( CuSO_4 + 5H_2O \to CuSO_4\cdot5H_2O \) (white to blue).

Teacher's Note:
1. The water vapour is produced when the compound is heated with CuO.

Answer: (c) simple distillation

Explanation:
1. The boiling points differ by 123 K, much greater than 80 K, so simple distillation is enough.

Teacher's Note:
1. Use fractional distillation only when the difference is small.

Answer: (b) fractional distillation

Explanation:
1. The difference in boiling points is only 9 K, so a fractionating column is required.

Teacher's Note:
1. Fractional distillation repeatedly condenses and vaporises the mixture.

Answer: (a) distillation under reduced pressure

Explanation:
1. Glycerol decomposes at its normal boiling point, so it is distilled at low pressure where it boils at a lower temperature.

Teacher's Note:
1. This method is also used to concentrate sugar juice.

Answer: (d) steam distillation

Explanation:
1. Aniline is immiscible with water and steam volatile, so it distils with steam below 373 K.

Teacher's Note:
1. Nitrobenzene and o-nitrophenol are also purified in this way.

Answer: (c) nitro compounds and azo compounds

Explanation:
1. Nitrogen in nitro, azo and azoxy groups and in rings such as pyridine is not completely converted to ammonium sulphate.

Teacher's Note:
1. Dumas' method is used for such compounds.

Answer: (b) nitrogen

Explanation:
1. The organic compound is heated with CuO; \( CO_2 \) is absorbed by KOH and the remaining nitrogen is collected.

Teacher's Note:
1. The oxides of nitrogen are reduced to N2 by hot copper gauze.