DPP-7 to 8 With Answer-English

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 7 Class : XIII Course : DPP No.1 Max. Time : 28 Total Time : 28 min. Single choice Objective ('–1' negative marking) Q.1 to Q.4 (3 marks 3 min.) [12, 12] Subjective Questions ('–1' negative marking) Q.5 to Q.8 (4 marks 4 min.) [16, 16] 1. 16 g of SOx occupies 5.6 litre at S.T.P. Assuming ideal gas nature, the value of x is (1) 1 (2*) 2 (3) 3 (4) none of these Sol. 5.6 𝑙 = 16 g 22.4 𝑙 = 16  22.4 = 64 5.6 gas 50 x as it M.W = 64 32 + x × 16 = 64  x = 2 Ans. 2. A and B are two elements which form AB2 and A2B3. If 0.2 mol of AB2 weighs 8 g and 0.1 mol of A2B3 weighs 7 g, the atomic weights of A & B will be : (1*) 20, 10 (2) 10, 10 (3) 15, 25 (4) 20, 20 Ans. A = 20, B = 10 Sol. 0.2 (MA + 2MB) = 8 0.1 (2MA + 3 MB) = 7 on solving MA = 20 ; MB = 10 3. Equal moles of H2O and NaCl are present in a solution. Hence, molality of NaCl solution is : (1) 0.55 (2*) 55.5 (3) 1.00 (4) 0.18 Sol. nH2O = nNaCl = n Mole of solute m = wt. of solvent(kg) = 1 n n 18 × 1000 = 18 ×1000 = 55.55 m. 4. When 5.08 g I2 and 3.55 g Cl2 are completely reacted according to the following reaction, calculate the number of moles of ICl & ICl3 produced. I2 + Cl2  ICl + ICl (1) 0.01 & 0.01 (2) 0.02 & 0.02 (3) 0.02 & 0.01 (4*) 0.01 & 0.03 5.08 Sol. POAC for I – 254 x 2 = x + y (1) POAC for Cl – 3.55 71 x 2 = x + 3y (2) by substracting eq. (1) from eq. (2) x + 3y = 0.05 x 2 (2) x + y = 0.02 x 2 (1) – – – 2y = 0.06 y = 0.03 Hence from eq. (1) x = 0.01 5. When 10 gm NaOH is added with 90 gm H2O gave a solution having density 1.2 g/ml then calculate for this solution. (i) % w/w (ii) % w/v (iii) mole fraction of NaOH Ans. (i) % w/w = 10% (ii) % w/v = 12% (iii) mole fraction of NaOH = 1/21 Sol. (i) % (w/w) = (ii) % (w/v) = 10 90  10 × 100 = 10% 10 (90  10) × 100 = 12% 1.2 (iii) Mole fraction of NaOH = 10 1 40 = 21 10  90 40 18 6. Calculate the molarity when (a) 4.9 g H2SO4 acid dissolved in water to result 500 ml solution (b) 2 gram molecule of KOH dissolved in water to result 500 ml solution. Ans. (a) 0.1 M (b) 4 M Sol. (a) M = 4.9 98 500 / 1000 2 mole (b) M = 500 × 1000 = 4 M. 7. What volume of 0.250 M HNO3 (nitric acid) reacts with 50 mL of 0.150 M Na2CO3 (sodium carbonate) in the following reaction ? Ans. 60 ml 2HNO3(aq) + Na2CO3(aq)  2NaNO3(aq) + H2O(𝑙) + CO2(g) Sol. 2HNO3(aq) + Na2CO3(aq)  2NaNO3(aq) + H2O(𝑙) + CO2(g) Millmole of HNO3 = 2 Millmole of Na2CO3 1 0.25 × Vml = 2 × 50 × 0.15 2  50  0.15 Vml = 0.25 = 60 ml. 8. Calculate individual and average Oxidation number (if required) of the marked element and also draw the structure of the following compounds or molecules. (6) S8 (12) CrO 2– (13) Cr2O 2– (14) Cr O Cl (15) CrO (16) Na H PO (17) FeS (18) C H O (19) NH4 NO3 NOTE : This question will not be discusse in class Ans. (1) +2 (6, –2) (2) +5/2(5, 5, 0, 0) (3) +6 (4) +6 (+6, +6) (5) +6 (+6, +6) (6) 0 (7) 0 (8) +5 (9) 4/3 (+ 2, +2, 0) (10) +8 (11) –3 (12) +6 (13) +6 (+6, +6) (14) +6 (15) +6 (16) +5 (17) +2 (18) 0 (19) –3, +5 Note : Inside the bracket ( ) answer for indivisual oxidation numbre is given. DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 8 Class : XIII Course : DPP No.2 Max. Marks : 27 Total Time : 27 min. Single choice Objective ('–1' negative marking) Q.1 (3 marks 3 min.) [3, 3] Subjective Questions ('–1' negative marking) Q.2 to Q.7 (4 marks 4 min.) [24, 24] 1. The vapour densities of two gases are in the ratio of 1 : 3 Their molecular masses are in the ratio of : (1*) 1 : 3 (2) 1 : 2 (3) 2 : 3 (4) 3 : 1 2. Calculate individual and average Oxidation number (if required) of the marked element and also draw the structure of the following compounds or molecules. (1) F N O3 (2) CaOCl2 (3) Xe O3 F2 (4) Li Al H4 (5) Na3 Al F6 (6) P4 (7) O3 (13) Si (OH)4 (8)  (O3)3 (14) Mg2 C3 (9) Fe3O4 (15) CaC2 (10) H2 SiF6 (16) Be2 C (11) P(OH)3 (12) POCl3 Ans. (1) +5 (2) 0 (–1, +1) (3) +8 (4) +3 (5) +3 (6) 0 (7) 0 8 (8) +3,+5 (9) 3 (+2, +3) (10) +4 (11) +3 (12) +5 (13) +4 4 (14) – 3 (–1, –1, –2) (15) – 1 (16) –4 Note : Inside the bracket ( ) answer for indivisual oxidation number is given. 3. Balance the reactions : (i) S2O3–2 + Sb2O5  SbO + H2SO3 Ans. 3S2O32– + 2Sb2O5 + 6H+ + 3H2O  4SbO + 6H2SO3 (ii) Cr2O7–2 + I– +H+  Cr+3 + I2 + H2O Ans. Cr2O72– + 6I– + 14H+  2Cr3+ + 3I2 + 7H2O (iii) IO4– + I– + H+  I2 + H2O Ans. IO4– + 7I– + 8H+  4I2 + 4H2O. 4. A sample contains 9.81 g Zn, 1.8 × 1023 atoms of Cr and 0.6 g–atoms of O. What is empirical formula of compound ? (Zn = 65 g/mol) Ans. ZnCr2O4. 9.81 Sol. Mole of zinc = 65 Mole of Cr = 1.8 1023 6.023 1023 = 0.15 = 0.3 Mole of O = 0.6 0.15 So simple ratio Zn = 0.15 Cr = 0.3 0.15 0.6 O = 0.15 Hence ZnCr2O4. = 1 = 2 = 4 5. What volume of water is required to make 0.20 M solution from 16 mL of 0.5 M solution ? Ans. 24ml 6. How many millilitres of 0.150 M H2SO4 (sulfuric acid) are required to react with 2.05 g of sodium hydrogen carbonate, NaHCO3, according to the following equation ? Ans. 81.3 mL H2SO4(aq) + 2NaHCO3(aq)  Na2SO4(aq) + 2H2O(l) + 2CO2(g) 7. A 6.90 M solution of KOH in water has 30% by weight of KOH. Calculate density of solution. Ans. 1.288 g/𝑙t

Comments

Post a Comment

Popular posts from this blog

Planning to start your own coaching institute for JEE, NEET, CBSE or Foundation? Here's why not to waste time in developing study material from scratch. Instead use the readymade study material to stay focused on quality of teaching as quality of teaching is the primary job in coaching profession while study material is secondary one. Quality of teaching determines results of your coaching that decides success & future of your coaching while good quality study material is merely a necessary teaching aid that supplements your teaching (video in Hindi)

Image
Read more

Physics-30.24-Physics-Solids and Semiconductors

UNIT 24 - SOLIDS AND SEMICONDUCTORS 1. SOLID STATE ELECTRONICS (SEMICONDUCTORS) (A) Energy bands in solids: (i) In solids, the group of closely lying energy levels is known as energy band. (ii) In solids the energy bands are analogous to energy levels in an atom. (iii) In solids the atoms are arranged very close to each other. In these atoms there are discrete energy levels of electrons. For the formation of crystal these atoms come close together, then due to nucleus-nucleus, electron-electron and electron-nucleus interactions the discrete energy levels of atom distort and consequently each energy level spits into a large number of closely lying energy levels. (iv) The number of split energy levels is proportional to the number of atoms interacting with each other. If two atoms interact then each energy level splits into two out of which one will be somewhat above and another will be somewhat below the main energy level. In solids the number of atoms is very large ( 1023). Hence eac...
Read more

Physics-31.Rotational Mechanics

5.1 DEFINITION OF CENTRE OF MASS Centre of mass: Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system. When the system moves under some external forces than this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at his point of translational motion. This point is called the centre of mass of the system. Centre of mass of system of n point masses or n particles is that point about which moment of mass of the system is zero, it means that if about a particular origin the moment of mass of system of n point masses is zero, then that particular origin is the centre of mass of the system. The concept of centre of mass is a pure mathematical concept. If there are n particles having mass m1, m2 ….m n and are placed in space (x1, y1, z1), (x2, y2, z2) ……….(x n, y n, z n) then centre of mass of system is defined as (X, Y, Z) where = Y = and Z = where...
Read more