Practice Questions

Q58.The coordination number of an atom in a body-centered cubic structure is______ [Assume that the lattice is made up of atoms.]

Q58. AB2 is 10% dissociated in water to A2+ and B− . The boiling point of 10. 0 molal aqueous solution of AB2 is-- ∘C. (Round off to the Nearest Integer). [Given : Molal elevation constant of water Kb = 0. 5 K kg mol−1 boiling point of pure water = 100°C]

Q58.The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is ______ ×10−5 mol dm−3. (Round off to the Nearest Integer). [Given : Henry's law constant = KH = 8. 0 × 104 kPa for O2. Density of water with dissolved oxygen = 1. 0 kg dm−3 ]

Q58.The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in kJmol−1 is ___ .(Rounded-off to the nearest integer) [R = 8. 314 J K−1 mol−1]

Q58.The spin-only magnetic moment value for the complex [Co(CN)6]4− is...... BM. (nearest integer value) [At. no. of Co = 27]

Q58.Assume a cell with the following reaction Cu(s) + 2 Ag+(1 × 10−3M) →Cu2+(0. 250M) E∘cell = 2. 97 V JEE Main 2021 (22 Jul Shift 1) JEE Main Previous Year Paper Ecell for the above reaction is V. (Nearest integer) [ Given : log 2. 5 = 0. 3979, T = 298 K]

Q58.If the conductivity of mercury at 0°C is 1 . 07 × 106 S m-1 and the resistance of a cell containing mercury is 0 . 243𝛺, then the cell constant of the cell is 𝑥× 104 m-1 . The value of 𝑥 is ______ . (Nearest integer)

Q58.A 5. 0 m mol dm−3 aqueous solution of KCl has a conductance of 0. 55 mS when measured in a cell constant 1. 3 cm−1 . The molar conductivity of this solution is _______ mSm2 mol−1 . (Round off to the Nearest Integer)

Q58.The following data was obtained for chemical reaction given below at 975 K. 2 NO(g) + 2H2( g) →N2( g) + 2H2O(g) JEE Main 2021 (26 Aug Shift 1) JEE Main Previous Year Paper [NO] [H2] Rate mol L−1 mol L−1 molL−1 s−1 (A) 8 × 10−5 8 × 10−5 7 × 10−9 (B) 24 × 10−5 8 × 10−5 2. 1 × 10−8 (C) 24 × 10−5 32 × 10−5 8. 4 × 10−8 The order of the reaction with respect to NO_____ is [Integer answer]

Q58.For the reaction A →B, the rate constant k (in s-1) is given by 2 . 47 × 103 log10k = 20 . 35 - T The energy of activation in kJ mol-1 is ________ . (Nearest integer) [Given : R = 8 . 314 J K-1 mol-1]

Q58.For the cell Cu(s) Cu2+(aq)(0. 1M)∥Ag+(aq)(0. 01M) Ag(s) the cell potential E1 = 0. 3095 V . For the cell Cu(s) Cu2+(aq)(0. 01M)∥Ag+(aq)(0. 001M) Ag(s) the cell potential = x × 10−2 V. Find value of x (Round off the Nearest Integer). [ Use : 2.303F RT = 0. 059 J ]

Q58.For the galvanic cell, Zn(s) + Cu2+(0. 02M) →Zn2+(0. 04M) + Cu(s) Ecell = … × 10−2 V (Nearest integer) : E0 Cu / Cu2+ = −0. 34 V, EZn / Zn2+ = +0. 76 V, 2.303F RT = 0. 059 V] [Use

Q58.When 3. 00 g of a substance X ' is dissolved in 100 g of CCl4 , it raises the boiling point by 0. 60 K. The molar mass of the substance ' X′ is ___ g mol−1 . (Nearest integer). [Given Kb for CCl4 is 5. 0 K kg mol−1]

Q58.An exothermic reaction X →Y has an activation energy 30 kJ mol−1 . If energy change ΔE during the reaction is −20 kJ, then the activation energy for the reverse reaction in kJ is (Integer answer) JEE Main 2021 (26 Feb Shift 1) JEE Main Previous Year Paper Q59. 3. 12 g of oxygen is adsorbed on 1. 2 g of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K in L is [R = 0. 0821 L atm K−1 mol−1]

Q58.The reaction 2 A + B2 →2 AB is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of _____ . (Round off to the Nearest Integer). JEE Main 2021 (17 Mar Shift 2) JEE Main Previous Year Paper

Q58.CO2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO2 exerts a partial pressure of 0 . 835 bar then x m mol of CO2 would dissolve in 0 . 9 L of water. The value of x is _______. (Nearest integer) (Henry's law constant for CO2 at 298 K is 1 . 67 × 103 bar)

Q58. PCl5(g) →PCl3(g) + Cl2(g) In the above first order reaction the concentration of PCl5 reduces from initial concentration 50 mol L−1 to 10 mol L−1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x × 10−2 min−1. The value of x is __________. [Given log 5 = 0. 6989 ]

Q58.Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3. 33 h at 25°C. After 9 h, the fraction of sucrose remaining is f. The value of log10( 1f ) is ________ ×10−2. (Rounded off to the nearest integer) [Assume: In 10 = 2. 303, ln 2 = 0. 693]

Q58.The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h2 . The value of x ma20 10 x is ( a0 is radius of Bohr's orbit) (Nearest integer) [Given: π = 3. 14]

Q58.Consider the sulphides HgS, PbS, CuS, Sb2 S3, As2 S3 and CdS. Number of these sulphides soluble in 50% HNO3 is ___________ .

Q58.The conductivity of a weak acid HA of concentration 0. 001 mol L−1 is 2. 0 × 10−5 S cm−1 . If Λ0m(HA) = 190 S cm2 mol−1, the ionization constant (Ka) of HA is equal to ____ ×10−6 (Round off to the Nearest Integer)

Q58.The resistance of conductivity cell with cell constant 1. 14 cm−1 , containing 0. 001M KCl at 298 K is 1500Ω . The molar conductivity of 0. 001M KCl solution at 298 K in S cm2 mol−1 is (Integer answer)

Q59.For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this reaction is ×10−3 s−1. (Round off to the Nearest Integer). [Given: log10 2 = 0. 301, ln 10 = 2. 303 ]

Q59.For coagulation of 50 mL of a sol, 10 mL of 0.5M Cl- ion solution is required. What is the coagulating value of Cl− ion solution (Nearest integer) NOTE: NTA question has been changed because it had errors.}

Q59.The first order rate constant for the decomposition of CaCO3 at 700 K is 6. 36 × 10−3 s−1 and activation energy is 209 kJ mol−1 . Its rate constant (in s−1) at 600 K is x × 10−6 . The value of x is (Nearest integer) [Given R = 8. 31 J K−1 mol−1; log 6. 36 × 10−3 = −2. 19, 10−4.79 = 1. 62 × 10−5]