Practice Questions
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Q53.If the solubility product of PbS is 8 Γ 10β28 , then the solubility of PbS in pure water at 298 K is x Γ 10β16 mol Lβ1 . The value of x is____ (Nearest integer) [Given β2 = 1. 41]
Q53.A rigid nitrogen tank stored inside a laboratory has a pressure of 30 atm at 06: 00 am when the temperature is 27 Β°C. At 03: 00pm, when the temperature is 45 Β°C, the pressure in the tank will be _____ atm. [nearest integer]
Q53.Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0. 55 g of the compound neutralised 12. 5 mL of 1M H2 SO4 solution. The percentage of nitrogen in the compound is (Nearest integer)
Q54. pH value of 0. 001 MNaOH solution is
Q54.A 2. 0 g sample containing MnO2 is treated with HCl liberating Cl2 . The Cl2 gas is passed into a solution of KI and 60. 0 mL of 0. 1 MNaS2 O3 is required to titrate the liberated iodine. The percentage of MnO2 in the sample is____. Nearest integer) [Atomic masses (in u) Mn = 55; Cl = 35. 5 : O = 16, I = 127, Na = 23, K = 39, S = 32 ]
Q54.At 600 K, 2 mol of NO are mixed with 1 mol of O2 . 2 NO(g) + O2(g) β2 NO2(g) The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0. 6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is____.
Q56.A sample of 0. 125 g of an organic compound when analysed by Duma's method yields 22. 78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mmHg . The percentage of nitrogen in the given organic compound is____ (a) The vapour pressure of water at 280 K is 14. 2 mmHg (b) R = 0. 082 L atm Kβ1 molβ1 JEE Main 2022 (28 Jul Shift 2) JEE Main Previous Year Paper
Q56.The resistance of a conductivity cell containing 0. 01 MKCl solution at 298 K is 1750 Ξ©. If the conductivity of 0. 01 MKCl solution at 298 K is 0. 152 Γ 10β3 S cmβ1 , then the cell constant of the conductivity cell is____ Γ10β3 cmβ1
Q57.In the presence of sunlight, benzene reacts with Cl2 to give product, X . The number of hydrogens in X is
Q57.The total number of monobromo derivatives formed by the alkanes with molecular formula C5H12 is (excluding stereo isomers)
Q58.The amount of charge in F (Faraday) required to obtain one mole of iron from Fe3 O4 is ______. (Round off the answer to the nearest integer)
Q58.(a) Baryte, (b) Galena, (c) Zinc blende and (d) Copper pyrites. How many of these minerals are sulphide based?
Q59.In the given reaction the number of sp2 hybridised carbon (s) in compound β²Xβ² is _______.
Q59.The difference in oxidation state of chromium in chromate and dichromate salts is___
Q59.The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is
Q59.The number of statement(s) correct from the following for Copper is/are JEE Main 2022 (27 Jun Shift 1) JEE Main Previous Year Paper (A) Cu(II) complexes are always paramagnetic (B) Cu(I) complexes are generally colourless (C) Cu(I) is easily oxidized (D) In Fehling solution, the active reagent has Cu(I)
Q59.For a first order reaction A βB, the rate constant, k = 5. 5 Γ 10β14 sβ1 . The time required for 67% completion of reaction is x Γ 10β1 times the half life of reaction. The value of x is Nearest integer) (Given : log 3 = 0. 4771 )
Q59.The conductivity of a solution of complex with formula CoCl3 (NH3)4 corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is
Q60.The number of oxygen present in a nucleotide formed from a base, that is present only in RNA is
Q60. In alanylglycylleucylalanylvaline the number of peptide linkages is:
Q60.How many of the given compounds will give a positive Biuret test____? Glycine, Glycylalanine, Tripeptide, Biuret
Q60.The spin only magnetic moment of the complex present in Fehling's reagent is____ B. M. (Round off your answer to the nearest integer) 1 2 5 5) 3 3) 3 + β5(log5 = 0 then
Q60.The number of chlorine atoms in bithionol is 1
Q61.The minimum value of the sum of the squares of the roots of π₯2 + 3 - ππ₯= 2π- 1 is (1) 6 (2) 4 (3) 5 (4) 8
Q61.Let the minimum value v0 of v = |z|2 + |z β3|2 + |z β6i|2 , z βC is attained at z = z0 . Then Β―2z20 βz30 + 3 2 + v20 is equal to (1) 1000 (2) 1024 (3) 1105 (4) 1196