Unit 13: Purification and Characterisation of Organic Compounds — Online MCQ Test
CHEMISTRY · JEE MAINS · JEE
Practice Unit 13: Purification and Characterisation of Organic Compounds with a free chapter-wise online MCQ test.
This chapter covers: Purification - Crystallization sublimation distillation differential extraction and chromatography-principles and their applications. Qualitative analysis - Detection of nitrogen s....
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Unit 13: Purification and Characterisation of Organic Compounds — Important Questions & Answers
When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL, 1M H2SO4 was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is:
- A. 21
- B. 0.42
- C. 42
- D. 0.21
Answer: C. 42
Eq of H2SO4 used = 15 * 1 * 2 = 30 milliequivalents. Since N-factor of NH3 is 1, moles of NH3 = 30 * 10^-3 = 0.03 mol. Mass of Nitrogen = 0.03 * 14 = 0.42g. Percentage = (0.42 / 1) * 100 = 42%.
Eq of H2SO4 used = 15 * 1 * 2 = 30 milliequivalents. Since N-factor of NH3 is 1, moles of NH3 = 30 * 10^-3 = 0.03 mol. Mass of Nitrogen = 0.03 * 14 = 0.42g. Percentage = (0.42 / 1) * 100 = 42%.
0.25 g of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of CaCl2 tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g , respectively. The percentage of oxygen in compound A is ____ %. (Nearest integer) (Given : molar mass in gmol-1 H :1, C :12, O :16)
- A. 73
- B. 75
- C. 68
- D. 80
Answer: A. 73
Mass of C = (12/44) * 0.1837 = 0.05 g. Mass of H = (2/18) * 0.15 = 0.0167 g. Mass of O = 0.25 - (0.05 + 0.0167) = 0.1833 g. %O = (0.1833/0.25)*100 = 73.32% ≈ 73%.
Mass of C = (12/44) * 0.1837 = 0.05 g. Mass of H = (2/18) * 0.15 = 0.0167 g. Mass of O = 0.25 - (0.05 + 0.0167) = 0.1833 g. %O = (0.1833/0.25)*100 = 73.32% ≈ 73%.
A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is (Molar mass S=32, BaSO4=233):
- A. 42.10%
- B. 63.15%
- C. 21.05%
- D. 48.24%
Answer: C. 21.05%
%S = (Atomic mass of S / Molar mass of BaSO4) * (Mass of BaSO4 formed / Mass of compound) * 100. %S = (32 / 233) * (0.4813 / 0.314) * 100 = 21.05%.
%S = (Atomic mass of S / Molar mass of BaSO4) * (Mass of BaSO4 formed / Mass of compound) * 100. %S = (32 / 233) * (0.4813 / 0.314) * 100 = 21.05%.
Sodium fusion extract of an organic compound (Y) with CHCl3 and chlorine water gives violet color to the CHCl3 layer. 0.15 g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is ____%. (Nearest integer). Molar masses: C=12, H=1, Cl=35.5, Br=80, I=127.
- A. 41
- B. 42
- C. 43
- D. 44
Answer: C. 43
The violet color indicates Iodine. Silver halide is AgI (molar mass = 108 + 127 = 235 g/mol). Mass of I = (Mass of AgI * Atomic weight of I) / Molar mass of AgI = (0.12 * 127) / 235 ≈ 0.06485 g. % of I = (Mass of I / Mass of sample) * 100 = (0.06485 / 0.15) * 100 ≈ 43.23%. Nearest integer is 43.
The violet color indicates Iodine. Silver halide is AgI (molar mass = 108 + 127 = 235 g/mol). Mass of I = (Mass of AgI * Atomic weight of I) / Molar mass of AgI = (0.12 * 127) / 235 ≈ 0.06485 g. % of I = (Mass of I / Mass of sample) * 100 = (0.06485 / 0.15) * 100 ≈ 43.23%. Nearest integer is 43.
Which one of the following is not an example of calcination?
- A. $\mathrm{Fe_2O_3\cdot xH_2O \xrightarrow{\Delta} Fe_2O_3 + xH_2O}$
- B. $\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}$
- C. $\mathrm{CaCO_3\cdot MgCO_3 \xrightarrow{\Delta} CaO + MgO + 2CO_2}$
- D. $\mathrm{2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2}$
Answer: D. $\mathrm{2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2}$
Calcination is heating an ore in the absence or limited supply of air to remove volatile matter, moisture, or to convert carbonate/hydrated ores into oxides. Options (A), (B), and (C) are calcination processes. Option (D) involves heating in the presence of oxygen, i.e. roasting of a sulphide ore, not calcination.
Calcination is heating an ore in the absence or limited supply of air to remove volatile matter, moisture, or to convert carbonate/hydrated ores into oxides. Options (A), (B), and (C) are calcination processes. Option (D) involves heating in the presence of oxygen, i.e. roasting of a sulphide ore, not calcination.