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The decomposition of a certain mass of CaCO3 gave 11.2 dm³ of CO₂ gas at STP. The mass of KOH required to completely neutralise the gas is:

A. 56 g B. 28 g C. 42 g D. 20 g Solution: (B) Weight of 11.2 dm3 of CO2 gas at STP = 44 / 2 KOH + CO2 → KHCO3 56g 44g KOH... View Article

The stoichiometry of the following reaction is:

K2S2O8(aq) + 2KI(aq) → 2K2SO4(aq) + I2(aq) A. 2 : 2 B. 1 : 1 C. 1 : 2 D. 2 : 1 Solution: (C) Generally, the stoichiometry of a... View Article

1 mole of methyl amine on reaction with nitrous acid gives at NTP

A. 1.0 L of nitrogen B. 22.4 L of nitrogen C. 11.2 L of nitrogen D. 5.6 L of nitrogen Solution: (B) The reaction: CH3−NH2 +... View Article

Sodium nitrate on reduction with Zn in presence of NaOH solution produces NH₃ Mass of sodium nitrate absorbing 1 mole of electron will be:

A. 7.750 B. 10.625 C. 8.000 D. 9.875 Solution: (B) Lets look at the following reaction: Zn + 2OH− ⟶ ZnO22− + 2H+ +... View Article

The ratio of amounts of H₂S needed to precipitate all the metal ions from 100mL of 1MAgNO₃ and 100mL of 1MCuSO₄ will be:

A. 1:1 B. 1:2 C. 2:1 D. None of these Solution: (B) 100 mL of 1 M AgNO3 = 0.1 mol AgNO3 100 mL of 1 M CuSO4 = 0.1 mol CuSO4 2AgNO3... View Article

Match the following Column I and Column II.

A. A - 4, B - 1, C - 2, D - 3 B. A - 5, B - 1, C - 2, D - 3 C. A - 4, B - 1, C -3, D - 2 D. A - 1, B - 4, C - 2, D - 3 Solution: (A) -... View Article

Acidified KMnO₄ oxidizes oxalic acid to CO₂. What is the volume (in litre) of 10⁻⁴MKMnO₄ required to completely oxidize 0.5litre of 10⁻²M oxalic acid in acid medium?

A. 125 B. 1250 C. 200 D. 20 Solution: (D)... View Article

Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1N Hydrochloric acid. The volume of Hydrochloric acid required is: (At. mass of calcium = 40)

A. 500 cm3 B. 400 cm3 C. 300 cm3 D. 200 cm3 Solution: (A) Let us find the volume of hydrochloric acid. No. of millimoles of Ca(OH)2 =... View Article

An aqueous solution containing 6.5gm of NaCl of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1M acetic acid required to neutralise NaOH obtained above is:

A. 100 cm3 B. 200 cm3 C. 1000 cm3 D. 2000 cm3 Solution: (A) Let us calculate the weight of pure NaCl = 6.5 × 0.9 = 5.85 g No.... View Article

A mixture of CaCl2 and NaCl weighing 4.44g is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56g of CaO. The percentage of NaCl in the mixture is:

(Atomic mass of Ca=40) A. 30.6 B. 75 C. 69.4 D. 25 Solution: (B) Weight of CaO = 0.56 g 0.56 g of CaO contains 0.40 g of calcium... View Article

Stoichiometric ratio of sodiumdihydrogen orthophosphate and sodium hydrogen orthophosphate required for synthesis of Na5P3O10 is:

A. 1.5:3 B. 3:1.5 C. 1:1 D. 2:3 Solution: (A)... View Article

The mass of potassium dichromate crystals required to oxidise 750 cm³ of 0.6 M Mohr’s salt solution is: (Given molar mass potassium dichromate = 294, Mohr’s salt = 392)

A. 0.45 g B. 22.05 g C. 2.2 g D. 0.49 g Solution: (B) Mohr's salt is FeSO4.(NH4)2S04.6H20 Only oxidisable part Fe2+ is;... View Article

5 moles of Ba(OH)₂ are treated with excess of CO₂. How much Ba(OH₂) will be formed?

A. 39.4 g B. 197 g C. 591 g D. 985g Solution: (D) Ba(OH)2 + CO2 → BaCO3 + H2O Given, 5 moles of barium hydroxide reacts with... View Article

If 20 g of CaCO₃ is treated with 100 mL of 20% HCl solution, the amount of CO₂ produced is:

A. 22.4 L B. 8.80 g C. 4.40g D. 2.24 L Solution: (B) CaCO3 + 2HCl → CaCl + CO2 + H2O 100 g 73 g 44 g 100mL of 20% HCl solution... View Article

A 5.82g silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as AgCl . The AgCl precipitate weighs 7.20g . The percentage of silver in the coin is;

A. 60.3% B. 80% C. 93.1% D. 70% Solution: (C) Weight of AgCl precipitated = 7.2 g Weight of Ag in the precipitate = (108 g) / (143.5... View Article

An organic compound has an empirical formula CH₂O, its vapour density is 45. The molecular formula of the compound is:

A. CH2O B. C2H5O C. C2H2O D. C3H6O3 Solution: () The empirical formula is CH2O. Therefore, the empirical weight = 30 g. Given, the... View Article

The empirical formula of a compound is CH₂. One mole of this compound has a mass of 42g. Its molecular formula is:

A. C3H6 B. C3H8 C. CH2 D. C2H2 Solution: (A) The empirical formula is CH2 Empirical weight of CH2 = 12 + (1 × 2) = 12 + 2 = 14... View Article

An organic compound contains 10.06 % carbon 0.84 % hydrogen and 89.10 % chlorine. What will be the empirical formula of the substance?

A. CH2Cl2 B. CCl4 C. CHCl3 D. CH3Cl Solution: (C) Carbon. 10.06 / 12 = 0.8383 Hydrogen 0.84 / 1 = 0.84 Chlorine. 89.1 / 35.5 = 2.51... View Article

The equivalent weight of a certain trivalent element is 20. The molecular weight of its oxide is;

A. 168 B. 68 C. 152 D. 56 Solution: (A) Atomic mass of element = Equivalent mass of element × Valency Atomic mass of element =... View Article

If molecular weight of KMnO₄ is M, then its equivalent weight in acidic medium would be;

A. M B. M / 2 C. M / 5 D. M / 3 Solution: (C) MnO−4 + 8H+ + 5e− → Mn2 + 4H2O Electrons gained = 5 Molecular weight... View Article