Chemistry-3.PART TEST-2-MAILING-Questions

PART TEST - II (MAILING) CHEMISTRY Time: Two Hours Maximum Marks : 100 Note: i) This paper has SEVEN questions. ii) Separate answers are to be given on the separate pages. iii) Attempt all questions. iv) Use of logarithmic table is PERMITTED v) Use of calculator is NOT PERMITTED Useful Data: Gas Constant R = 8.314 J mol−1 K−1 = 0.0821 lit atm mol−1 K−1 = 2 Cal mol−1 Avogadro's Number Na = 6.023 × 1023 \Planck’s constant h = 6.625 × 10−34 J sec. 1 Faraday = 96500 Coulomb 1 calorie = 4.2 Joule Atomic No: Ca = 20, C = 6, O = 8, K = 19, Cl = 17, N = 7, S = 16, Na = 11, Cu = 29, Co =27, Mn = 25. Atomic Masses: Ag =108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39, Cl = 35.5, N = 14, S = 32, Na = 23, H = 1, I = 127, Mg = 24, Ne = 20, Br = 80, I = 127, As = 75, Cu = 63.54 Name : ☐☐☐☐☐☐☐☐☐☐☐☐☐☐☐☐☐ Enrol No. : ☐☐☐☐☐☐☐☐☐☐☐☐☐ 1. a) In an electrolysis experiment, current was passed for 5 hours through two cells connected in series. The first cell contain a solution of gold and the second contains a CuSO4 solution 9.58 gm of gold was deposited in the first cell. If oxidation no of gold is +3, find the amount of Cu deposited on cathode in second cell. Also calculate the current strength in ampere [8] b) The standard reduction potential of Cu2+/Cu and Ag+/Ag electrodes are 0.337 V and 0.799 V respectively. Construct a galvanic cell using these electrodes so that its E°cell is +ve. For what [Ag+] will the e.m.f of the cell at 25°C be zero if [Cu++] is 0.01 M [8] 2. a) Calculate the heat of dissociation of N2O4 into NO2 from the data N2(g) + O2(g) → 2NO(g) ΔH = + 43.10 Kcal N2(g) + 2O2(g) → N2O4(g) ΔH = - 1.87 Kcal 2NO2(g) → 2NO(g) → O2(g) ΔH = + 26.10 Kcal [8] b) A solution of 5 gm of hemoglobin (mol.wt.64000) in 100 cc of solution shows a temperature rise of 0.31°C for complete oxygenation. Each mole of hemoglobin binds 4 moles of oxygen. If the heat capacity of the solution is 4.18 JK-1 cm-3, Calculate ΔH per gm mole of oxygen bound. [8] 3. a) Compute the heat of formation of liquid methanol in KJ/mol, using the following data. Heat of vaporization of liquid methanol =38 KJ/mol. Heat of formation of gaseous atoms from the element in their standard states H, 218 KJ/mol, C, 715 KJ/mol, O, 249 KJ/mo. [8] b) After electrolysis of a sodium chloride solution with inert electrodes for a certain period of time, 600 ml of the 1N solution was left which was found to be NaOH. During the same 31.75 gm copper was deposited in copper voltameter in series with the electrolytic cell. Calculate the percentage of NaOH obtained. [8] 4. a) Calculate the amount of ice that will separate out on cooling a solution containing 50 gm of ethylene glycol in 200 gm water to -9.3°C. [8] (Kf for water = 1.86 K mole-1 kg) b) Benzene and toluene form two ideal solutions A and B at 313 K. Solution A contains 4 moles of toluene and one mole of benzene. Solution B contain equal masses of toluene and benzene. Calculate total pressure in each case. The vapour pressure of benzene and toluene are 160 and 60 mm respectively at 313 K. [8] 5. a) Silver has an atomic radius of 144 pm. If density of silver is 10.6 g/cc, to which type of cubic crystal silver belongs ? [5] b) Give IUPAC names of the following complexes (i) Na2[Co(H2O)2(OH)4] (ii) [Co(en)2 (NO2) (SCN)]Cl [5] 6. a) An orange solid (A) on heating gives a green residue (B), a colorless gas (C) and water vapour. the dry gas (C) on passing over heated Mg gave a solid (White) (D). (D) on reaction with water gave a gas (E) which formed dense white fumes with HCl. Identify (A) to (E) giving reaction. [6] b) Complete the following reaction : i) AgBr + Na2S2O3 → ii) P4 + HNO3 → iii) Pb3O4 + HNO3 → [6] 7. a) The edge length of a cubic unit cell of metallic calcium is 556 pm. If the density of calcium is 1.54 gm/cc and its molar concentration is 40.08 gm/mol, calculate the radius of calcium atom. [6] b) An unknown solid mixture contains one or two of the following CaCO3, BaCl2, AgNO3,Na2SO4,ZnSO4 and NaOH. The mixture is completely soluble in water and the solution gives pink colour with phenolphthalein. When dil HCl is gradually added to the above solution, a precipitate is formed which dissolves with further addition of the acid. What is/are present in the solid ? Give equation to explain the appearance of the precipitate and its dissociation. [8] 6

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45º) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more