DPP-15 to 16 With Answer Physical Chemistry

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 15 Class : XIII Course : DPP No.1 Max. Time : 31 Total Time : 31 min. Single choice Objective ('–1' negative marking) Q.1 to Q.9 (3 marks 3 min.) [27, 27] Subjective Questions ('–1' negative marking) Q.10 (4 marks 4 min.) [4, 4] 1. Equivalent weight of NH3 in the change N2  NH3 is : (1) 17/6 (2) 17 (3) 17/2 (4*) 17/3 2. A sample of 1.0 g of calcite (pure CaCO3) required 39.5 mL of HCl complete reaction. Calculate the normality of the acid. (1) 0.41 N (2) 0.61 N (3*) 0.51 N (4) 0.15 N 3. Number of moles of electrons taken up when 1 mole of NO  ions is reduced to 1 mole of NH OH is : (1) 2 (2) 4 (3) 5 (4*) 6 4. In the following reaction hydrazine is oxidized to N2. N H + OH–  N + H O + e The equivalent weight of N2H4 (hydrazine) is (1*) 8 (2) 16 (3) 32 (4) 64 5. Which of the following sets of quantum numbers can be correct for an electron in 4f-orbitals ? (1) n = 3, 𝑙 = 2, m = –2, s = + 1 (2) n = 4, 𝑙 = 4, m = –4, s = – 1 2 2 (3*) n = 4, 𝑙 = 3, m = +1, s = + 1 (4) n = 4, 𝑙 = 3, m = +4, s = + 1 2 2 Sol. For n = 4, 𝑙  4, for 𝑙 = 3, m  4 (n = 4, 𝑙  4, for 𝑙 = 3, m  4 ds fy,½ 6. In the following electronic configuration, some rules have been violated : I : Hund II : Pauli's exclusion III : Aufbau (1) I and II (2) I and III (3) II and III (4*) I, II and III 7. In the electronic configuration of Mn(Z = 25) : (1) the number of electrons having n + 𝑙 = 4 are 5 (2*) the number of electrons having m = 0 are 13 (3) the magnetic moment is 1.73 BM (4) Mn belongs to IIIrd period and d-Block in periodic table. Sol. Mn - 1s2 2s2 2p6 3s2 3p6 3d5 4s2 m = 0 for all s-electron and two-p electron and one d-electron hence total = 13. 8. A wavelength of 400 nm of electro magnetic radiation of corresponds to : (1) frequency () = 7.5 × 1014 Hz (2) wave number(  ) = 2.5 × 106 m–1. (3) momentum (m) = 1.66 × 10–27 kg ms–1 (4*) all are correct values. Sol. Use C =   u  1  9. The total volume of 0.1 M KMnO4 solution that are needed to oxidize 144 mg of ferrous oxalate and 152 mg of ferrous sulphate in a mixture in acidic medium is : (1) 5 mL (2) 2 mL (3*) 8 mL (4) None of these Sol. m-eq. of KMnO4 = m.eq. of FeC2O4 + m-eq. of FeSO4 (0.1 × 5) × V = 144 144 3 152 + 152 ; 1 V = 8 mL 10. Calculate ratio of wavelength for an proton and -particle if they are accelerated through same potential V : Ans. p  2 2  1 Sol. Use  = FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 16 Class : XIII Course : DPP No.2 Max. Marks : 30 Total Time : 30 min. Single choice Objective ('–1' negative marking) Q.1 to Q.10 (3 marks 3 min.) [30, 30] 1. Alveoli are the tiny sacs of air in the lungs whose average diameter is 50 pm. Consider an oxygen molecule trapped within a sac. Calculate uncertainty in the velocity of oxygen molecule ? (1) 1.98 × 10–2 ms–1 (2*) 19.8 ms–1 (3) 198 × 10–4 ms–1 (4) 19.8 × 10–6ms–1 Sol. x = 50 pm So v = h 4.mx 6.625  10 34  6.022  1023 = 4  3.14  32  103  50  1012 m/sec = 0.0019853 × 104 m/sec = 19.853 m/sec. 2. Magnetic moment of 25Mnx is B.M then the value of x is : (1) 1 (2) 2 (3) 3 (4*) 4 Sol. 25Mn – [Ar] 3d54s2 Given = 15  n = 3 Hence to have ‘3’ unpaired electrons Mn must be in ‘+4’ state. 3. The maximum number of electrons in s, p and d-subshells are : (1) 2 in each (2) 2, 6 and 6 (3*) 2, 6 and 10 (4) 2, 6 and 12 4. In the reaction 2CuSO + 4K  Cu  +  + 2K SO the equivalent weight of CuSO4 will be (atomic weight of Cu = 63.5) 5. The equivalent weight of a metal is double than of oxygen. How many times is the weight of it’s oxide greater than the weight of the metal ? (1*) 1.5 (2) 2 (3) 3 (4) 4 6. A 10-volume H2O2 solution is equal to (1) 3% (w/v) H2O2 (2) 30 g/L H2O2 (3) 1.76 N (4*) all of these mijksDr lHkh 7. 4 mole of a mixture of Mohr’s salt and Fe2(SO4)3 requires 500 mL 1M K2Cr2O7 for complete oxidation in acidic medium. The mole % of the Mohr’s salt in the mixture is : (1) 25 (2) 50 (3) 60 (4*) 75 Sol. Cr O 2– + 6Fe2+ (n = 1) (Mohr’s salt) + 14H+  2Cr3+ + 6Fe3+ + 7H O 2 7 2 Equivalent of Fe2+ = moles of Mohr’s salt = equivalent of K2Cr2O7 = 500 × 10–3 × 6 × 1 = 3.0 Hence, mole percent of Mohr’s salt = 3 × 100 = 75 4 8. Which one is the correct order of the size of the iodine species (1) I > I+ > I– (2) I > I– > I+ (3) I+ > I– > I (4*) I– > I > I+ 9. Radii of fluorine and neon in angstrom units are respectively given by (1*) 0.762, 1.60 (2) 1.60, 1.60 (3) 0.72, 0.72 (4) None of these values (1*) 0.762, 1.60 (2) 1.60, 1.60 (3) 0.72, 0.72 (4)buesalsdksbZekuugha Sol. Nobel gases have largest radii in the periodic table. 10. X2+ is isoelectronic of sulphur and has (Z + 2) neutrons (Z is atomic no. of X2+). Hence mass number of X2+ is : (1) 34 (2) 36 (3*) 38 (4) 40

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 45ΒΊ) 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