Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question One litre buffer solution was prepared by adding 0.10 mol each of NH₃ and NH₄Cl in deionised water. What is the change in pH on addition of 0.05 mol of HCl to the above solution? (Treat the solution volume change on addition as negligible — 1 L final volume.) 🖼️ Question Image ✍️ Short Solution (idea) This is a classical buffer problem . Use the Henderson–Hasselbalch equation for the NH₄⁺/NH₃ buffer: pH = p K a + log ⁡ [ base ] [ acid ]​ For the NH₄⁺/NH₃ system, p K a = 14 − p K b ( NH 3 ) \text{p}K_a = 14 - \text{p}K_b(\text{NH}_3) . Use p K b ( NH 3 ) = 4.75 \text{p}K_b(\text{NH}_3)=4.75  ⇒ p K a ( NH 4 + ) = 9.25 \text{p}K_a(\text{NH}_4^+)=9.25 . Initial moles (in 1 L): n NH 3 = 0.10 →  [ b a s e ] = 0.10   M [{\rm base}]=0.10\ \mathrm{M} n NH 4 + = 0.10 n_{\text{NH}_4^+}=0.10  → [ a c i d ] = 0.10   M [{\rm acid}]=0.10\ \mathrm{M} So initial pH: pH initial = 9.25 + log ⁡ 0.10 0.10 = 9.25 + 0 = 9.25. Add 0.05 mo...

  ❓ Question In Dumas' method , 292 mg of an organic compound released 50.0 mL of nitrogen gas (N₂) at 300 K and 715 mm Hg total pressure. Aqueous tension at 300 K = 15 mm Hg . Find the percentage composition of nitrogen (N) in the organic compound. (Nearest integer) 🖼️ Question Image ✍️ Short Solution Step 1 — Correct the gas pressure for water vapor Total pressure above the collected gas includes water vapor. The partial pressure of dry N₂ is: P N 2 = P total − P H 2 O = 715 − 15 = 700  mm Hg . Step 2 — Use ideal gas law to find moles of N₂ Use P V = n R T. Convert volume to litres:  V = 50.0  mL = 0.0500  L V = 50.0\ \text{mL} = 0.0500\ \text{L} . Use gas constant in mmHg·L units: R = 62.3637   L \ mmHg \ mol − 1 \ K − 1 . So n N 2 = P N 2   V R   T = 700 × 0.0500 62.3637 × 300 . Calculate numerator: 700 × 0.0500 = 35.0 700\times0.0500 = 35.0 . Denominator: 62.3637 × 300 = 18709.11 62.3637\times300 = 18709.11 . Thus n N 2 = 35...

❓ Question 1 M aqueous solutions of each of C u ( N O 3 ) 2 ,   A g N O 3 ,   H g 2 ( N O 3 ) 2 ,   M g ( N O 3 ) 2 \mathrm{Cu(NO_3)_2},\ \mathrm{AgNO_3},\ \mathrm{Hg_2(NO_3)_2},\ \mathrm{Mg(NO_3)_2}  are electrolysed using inert electrodes . Statement (I): With increasing (more negative) cathode potential (i.e. increasing applied voltage), the sequence of deposition of metals on the cathode will be Ag, Hg and Cu . Statement (II): Magnesium will not be deposited at cathode; instead oxygen gas will be evolved at the cathode. Decide which statement(s) are correct and explain. 🖼️ Question Image ✍️ Short Solution Key idea: Which species is reduced at the cathode depends on standard (or actual) reduction potentials. The species with the most positive reduction potential is reduced first (at the least negative cathode potential). As the applied cathode potential is driven more negative, species with lower reduction potentials begin to reduce. 🔹 Standard reduction pot...

  ❓ Question Which of the following statements are correct ? (A) Tl³⁺ is a powerful oxidising agent (B) Al³⁺ does not get reduced easily (C) Both Al³⁺ and Tl³⁺ are very stable in solution (D) Tl⁺ is more stable than Tl³⁺ (E) Al³⁺ and Tl⁺ are highly stable 🖼️ Question Image ✍️ Short Solution We evaluate each statement using periodic trends, standard reduction potentials, and the inert-pair effect. (A) Tl³⁺ is a powerful oxidising agent — True Thallium in +3 state is relatively unstable because of the inert-pair effect (the 6s electrons are reluctant to participate in bonding). Tl³⁺ tends to accept electrons to form the more stable Tl⁺ (6s²), so it oxidizes other species and itself gets reduced → hence Tl³⁺ acts as a strong oxidant in solution. (B) Al³⁺ does not get reduced easily — True Aluminium’s standard reduction potential (Al³⁺/Al) is strongly negative (≈ −1.66 V), meaning reduction to metallic Al is thermodynamically unfavorable in aqueous solution. The...

  ❓ Question Butane reacts with oxygen to produce carbon dioxide and water following the equation: C 4 H 10 ( g ) + 13 2   O 2 ( g ) ⟶ 4   C O 2 ( g ) + 5   H 2 O ( l ) If 174.0 kg of butane is mixed with 320.0 kg of O₂, the volume of liquid water formed (in liters, nearest integer) is ________. 🖼️ Question Image ✍️ Short Solution We solve by: Converting given masses to moles. Identifying the limiting reagent using reaction stoichiometry. Using stoichiometry to find moles (and mass) of H₂O produced. Converting mass of liquid water to volume (density ≈ 1.00 g·mL⁻¹). Step 1 — Molar masses & moles Molar mass (approx): M C 4 H 10 = 4 ( 12.01 ) + 10 ( 1.008 ) = 58.12   g \mol − 1 M_{\mathrm{C_4H_{10}}}=4(12.01)+10(1.008)=58.12\ \text{g·mol}^{-1} M O 2 = 32.00   g \ mol − 1 M_{\mathrm{O_2}}=32.00\ \text{g·mol}^{-1} M H 2 O = 18.015   g \ mol − 1 M_{\mathrm{H_2O}}=18.015\ \text{g·mol}^{-1} Moles of butane: n C 4 H 10 = 174000  g...

❓ Question Which of the following statements are correct ? (A) Tl³⁺ is a powerful oxidising agent (B) Al³⁺ does not get reduced easily (C) Both Al³⁺ and Tl³⁺ are very stable in solution (D) Tl⁺ is more stable than Tl³⁺ (E) Al³⁺ and Tl⁺ are highly stable 🖼️ Question Image ✍️ Short Solution We evaluate each statement using electronic structure, standard electrode potentials and chemical trends (inert-pair effect, stability of Al³⁺). 🔹 (A) Tl³⁺ is a powerful oxidising agent — True Tl in +3 state has electronic configuration [Xe]4f¹⁴5d¹⁰ (6s⁰6p¹ for neutral Tl → Tl³⁺ has lost the 6p and 6s electrons partially). Due to the inert-pair effect , the 6s electrons are reluctant to participate in bonding; Tl³⁺ is stabilized poorly and tends to be reduced to Tl⁺. Thus Tl³⁺ accepts electrons readily → it is a strong oxidizing agent (gets reduced to the more stable Tl⁺). Example: Tl³⁺ is often reduced in aqueous solution; TlCl₃ disproportionates, etc. 🔹 (B) Al³⁺ d...

❓ Question ‘ X ’ is the number of acidic oxides among: VO₂, V₂O₃, CrO₃, V₂O₅, Mn₂O₇ . The primary valency of cobalt in the complex [ C o ( H 2 N C H 2 C H 2 N H 2 ) 3 ] 2 ( S O 4 ) 3 [ \mathrm{Co}(\mathrm{H_2NCH_2CH_2NH_2})_3 ]_2(\mathrm{SO_4})_3 ​ (i.e. [ C o ( en ) 3 ] 2 ( S O 4 ) 3 [Co(\text{en})_3]_2(SO_4)_3 ​ ) is ‘ Y ’. Find the value of X + Y . 🖼️ Question Image ✍️ Short Solution We solve in two parts — first count acidic oxides (X), then find the primary valency (Y). 🔹 Part A — Identify acidic oxides (find X) For each oxide determine the oxidation state of the metal and typical acid/base character: VO₂ — V oxidation state: +4. Oxides of transition metals at intermediate oxidation states (like +2, +3, +4) are typically basic or amphoteric , not strongly acidic. VO₂ is amphoteric/neutral → not counted as acidic. V₂O₃ — V oxidation state: +3. Lower oxidation oxides are generally basic (or amphoteric) → not acidic . CrO₃ — Cr oxidation state: +...

  ❓ Question Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If xA and xB are the mole fraction of A and B in solution while yA and yB are the mole fraction of A and B in vapour phase then 🖼️ Question Image ✍️ Short Solution For an ideal solution Raoult’s law applies: p A    =    x A   p A ∗ , p B    =    x B   p B ∗ p_A \;=\; x_A\,p_A^*,\qquad p_B \;=\; x_B\,p_B^* ​ Total vapour pressure above the solution: p tot    =    p A + p B    =    x A p A ∗ + x B p B ∗ . p_{\text{tot}} \;=\; p_A + p_B \;=\; x_A p_A^* + x_B p_B^*. The mole fractions in the vapour (via Dalton’s law): y A    =    p A p tot = x A p A ∗ x A p A ∗ + x B p B ∗ , y_A \;=\; \frac{p_A}{p_{\text{tot}}} =\frac{x_A p_A^*}{x_A p_A^* + x_B p_B^*}, y B    =    p B p tot = x B p B ∗ x A p A ∗ + x B p B ∗ . y_B \;=\; \frac{p_B}{p_{\text{tot}}} =\frac{x_B p_B^*}{x_A p_A^* + x_B p_B^*}. Since x B = 1 − x A x_B=1-x_A ​ ...