Questions & explanations
1. A student performs flame tests on two salts. Salt X gives a brick-red flame; salt Y gives an apple-green flame. Which confirmatory test distinguishes X from Y?
- Add dilute H2SO4 to each; X gives a white precipitate, Y does not.
- Add ammonium oxalate solution; X gives a white precipitate insoluble in acetic acid, Y gives a white precipitate soluble in acetic acid.
- Add sodium cobaltinitrite; X gives a yellow precipitate, Y does not.
- Add NaOH and heat; X gives a gas that turns red litmus blue, Y does not.
Answer: Add ammonium oxalate solution; X gives a white precipitate insoluble in acetic acid, Y gives a white precipitate soluble in acetic acid.
Brick-red flame indicates Ca2+; apple-green indicates Ba2+. Ammonium oxalate gives CaC2O4 (white, insoluble in acetic acid) and BaC2O4 (white, soluble in acetic acid). This distinguishes them. Dilute H2SO4 gives BaSO4 (insoluble) with Ba2+, but CaSO4 is slightly soluble, so both may give precipitate. Sodium cobaltinitrite tests for K+, not Ca2+/Ba2+. NaOH+heat tests for NH4+.
2. An alloy of sodium and potassium is used as a coolant in fast breeder nuclear reactors. Which of the following statements about this alloy is correct?
- It is a solid at room temperature and contains 78% Na and 22% K.
- It is a liquid at room temperature and contains 78% Na and 22% K.
- It is a liquid at room temperature and contains 78% K and 22% Na.
- It is a solid at room temperature and contains 78% K and 22% Na.
Answer: It is a liquid at room temperature and contains 78% K and 22% Na.
The Na-K alloy forms a eutectic mixture with a melting point of -12.6°C, making it liquid at room temperature. The eutectic composition is approximately 78% K and 22% Na by weight. This low-melting liquid alloy is used as a coolant in fast breeder reactors due to its excellent heat transfer properties.
3. Cesium and potassium exhibit photoelectric effect with visible light, but lithium and sodium do not. What is the primary reason for this difference?
- Cesium and potassium have lower ionization enthalpies than lithium and sodium.
- Cesium and potassium have higher ionization enthalpies than lithium and sodium.
- Cesium and potassium have larger atomic radii than lithium and sodium.
- Cesium and potassium have smaller atomic radii than lithium and sodium.
Answer: Cesium and potassium have lower ionization enthalpies than lithium and sodium.
The photoelectric effect requires photons with energy greater than the work function of the metal. Cesium and potassium have low ionization enthalpies, hence low work functions, so visible light photons can eject electrons. Lithium and sodium have higher ionization enthalpies and require UV light.
4. Which alkali metal is the strongest reducing agent in aqueous solution?
- Cs
- Na
- Li
- K
Answer: Li
Reducing strength in aqueous solution is determined by standard electrode potential (E°). Li has the most negative E° (-3.04 V) among alkali metals due to its exceptionally high hydration enthalpy, which overcomes its high ionization energy. Thus, Li is the strongest reducing agent in water.
5. A water sample contains 0.162 g of Ca(HCO3)2 and 0.095 g of MgCl2 per litre. What is the total hardness in ppm of CaCO3 equivalent?
- 250 ppm
- 100 ppm
- 150 ppm
- 200 ppm
Answer: 200 ppm
Calculate CaCO3 equivalent: For Ca(HCO3)2, molar mass = 162 g/mol, CaCO3 equivalent mass = 100 g/mol. Mass of CaCO3 equivalent = (0.162/162)*100 = 0.100 g = 100 ppm. For MgCl2, molar mass = 95 g/mol, CaCO3 equivalent = 100 g/mol. Mass = (0.095/95)*100 = 0.100 g = 100 ppm. Total = 200 ppm.
6. In the electrolytic extraction of magnesium from fused MgCl2, which of the following is added to the electrolyte to lower its melting point?
- NaCl and CaCl2
- KCl and CaCl2
- NaCl and KCl
- CaCl2 and MgO
Answer: KCl and CaCl2
In the electrolytic extraction of magnesium, anhydrous MgCl2 is mixed with KCl and CaCl2 to lower the melting point of the electrolyte to about 700°C. This is the standard industrial practice as per NCERT Class 11 Chemistry, Chapter 10, Section 10.6.
7. Which of the following bicarbonates can be isolated as a solid?
- LiHCO3
- Ca(HCO3)2
- NaHCO3
- Mg(HCO3)2
Answer: NaHCO3
Among s-block bicarbonates, only NaHCO3 and KHCO3 can be isolated as solids. LiHCO3 and all alkaline earth bicarbonates exist only in solution because the small cation polarises the bicarbonate ion, causing decomposition upon attempted isolation.
8. Which of the following alkali metal halides has the highest melting point?
- NaCl
- LiF
- KBr
- CsI
Answer: LiF
Melting point of ionic compounds depends on lattice energy. LiF has the highest lattice energy due to small size of Li+ and F- ions, leading to strong electrostatic attraction. Hence, it has the highest melting point among the given halides.
9. In the Down's process for sodium extraction, what is the purpose of adding CaCl2 to NaCl before electrolysis?
- To lower the melting point of NaCl from 801°C to about 600°C
- To increase the electrical conductivity of the melt
- To prevent the formation of sodium oxide
- To act as a reducing agent for Na+ ions
Answer: To lower the melting point of NaCl from 801°C to about 600°C
In the Down's process, CaCl2 is added to NaCl to lower the melting point of the electrolyte from 801°C to about 600°C, reducing energy consumption. This is a standard fact from NCERT Class 11 Chemistry, Chapter 10, Section 10.1.
10. Which gas is passed through brine to obtain pure sodium chloride by crystallisation?
- Hydrogen chloride gas
- Chlorine gas
- Carbon dioxide gas
- Ammonia gas
Answer: Hydrogen chloride gas
The common-ion effect is used: passing HCl gas through brine provides Cl⁻ ions, which shifts the equilibrium NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq) to the left, causing pure NaCl to crystallise out while impurities remain in solution.
11. A water sample has both Ca(HCO3)2 and MgSO4. Which single method will remove both temporary and permanent hardness?
- Ion-exchange resin
- Clark's process
- Boiling
- Washing soda treatment
Answer: Ion-exchange resin
Ion-exchange resins remove both Ca2+ and Mg2+ ions by exchanging them with Na+ or H+ ions, thus eliminating both temporary and permanent hardness in a single step. This method is effective for all hardness-causing cations.
12. In self-contained breathing apparatus, KO₂ removes CO₂ and produces O₂. What mass of CO₂ (in g) is removed per gram of KO₂? (Molar masses: K=39, O=16, C=12)
- 0.62
- 0.31
- 0.15
- 0.93
Answer: 0.62
Reaction: 4KO₂ + 2CO₂ → 2K₂CO₃ + 3O₂. Molar mass KO₂ = 71 g/mol. 1 g KO₂ = 1/71 mol. From stoichiometry, 4 mol KO₂ remove 2 mol CO₂, so moles CO₂ removed = (2/4)*(1/71) = 1/142 mol. Mass CO₂ = (1/142)*44 = 44/142 ≈ 0.62 g.