DPP-45 to 46 English Physical PC

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 45 to 46 Class : XIII Course : DPP No.45 DPP No.1 1. The ionic product of water at 45 ºC is 4  1014. What is pH of pure water at this temperature. [Take : log 2 = 0.3] (1) 6.7 (2) 7 (3) 7.3 (4) 13.4 2. For which temperature the pKw of pure water can be greater than 14. (1) 20 ºC (2) 30 ºC (3) 40 ºC (4) 50 ºC 3. For pure water at 10 ºC and 60 ºC , the correct statement is (1) pOH10ºC = pOH60ºC (2) pOH10ºC > pOH60ºC (3) pOH60ºC > pOH10ºC (4)Can't say 4. For pure water at 25 ºC and 50 ºC the correct statement is (1) pH25ºC = pH50ºC (2) pH25ºC > pH50ºC (3) pH50ºC > pH25ºC (4) Can't say Passage : (Q.5 to Q.9) Relative strengths of conjugate acid base pairs : HCIO4  CIO4 –  very weak bases HCI   Strong acids CI–  Negligible tendency Stronger acid H2SO4  100% dissociated in aqueous solution HSO4 –  to be protonated in Weaker base HNO3  NO3 –  aqueous solution H O+ H O 3 2 HSO  H3PO4  SO2    HNO2  HF   H2PO4  NO  F  CH3COOH  H2CO3  CH3CO  Weak bases H2S NH  Weak acids  Exist in solution as  a mixture of HA HCO   HS   Moderate tendency to be protonated HCN  HCO   H2O  A ,and H3O NH3 CN CO2 OH  in aqueous solution       NH3    Very Weak acids NH  2  Strong bases 100% Weaker acid OH  Negligibletendency O2  protonated in Strong base 2  to dissociate.   H  aqueous solution 5. Account for the acidic properties of nitrous acid in terms of (i) Arrhenius theory and (ii) Bronsted Lowry theory 6. Write a balanced equation for the dissociation of each of the following Bronsted Lowry acids in water. (1) H SO (2) H O+ (3) HSO – 2 4 3 4 Also write conjugate base of the acid. 7. Which of the following reactions proceeds to the right and which proceeds to the left if you mix equal concentrations of reactants and products ? (1) HF(aq) + NO –(aq) HNO (aq) + F– (aq) (2) NH + (aq) + CO 2– (aq) HCO –(aq) + NH (aq) 3 3 4 3 3 3 8. Which of the following species behave as a strong acids or as strong base in aqueous solutions ? (a) HNO (b) HNO (c) NH + (d) Cl– (e) H– (f) O2– (g) H SO 2 3 4 2 4 9. Consider following reactions : (1) H CO (aq) + HSO– (aq) H SO (aq) + HCO – (aq) 2 3 4 2 4 3 (2) HF (aq) + Cl¯ (aq) HCl (aq) + F¯ (aq) (3) HF (aq) + NH (aq) NH + + F¯ (aq) (4) HSO – (aq) + CN¯ (aq) HCN (aq) + SO2– (aq) Reactions proceeding to the right are : (1) a, b (2) c, d (3) a, c (4) b, d Integer Answer Type This section contains 2 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. 10. The equilibrium N2(g) + 3H2(g) 2NH3(g) is established in a closed container by initially taking only NH3. Find out the number of moles of H2 present in 102 g of the equilibrium mixture if the molar mass of the equilibrium mixture is observed to be 51 g/mole. 4 11. For the following equilibrium PCl5(g) PCl3(g) + Cl2(g), vapour density at equilibrium is found to be 100. Initially 1 mole of PCl5 is taken in 12.82 lt. flask at 27ºC. Calculate equilibrium pressure (in atm) of the system. DPP No.46 DPP No.2 for week (03-01-11 to 08-01-11) Total Marks : 50 Max. Time : 50 min. Single choice Objective (no negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Subjective Questisons (no negative marking) Q.7 to Q.14 (4 marks 4 min.) [32, 32] 1. At – 50ºC autoprotolysis of NH gives [NH+ ] = 1 × 10–15 M hence, autoprotolysis constant of NH is: (1) (2) 1 × 10 –30 (3) 1 × 10–15 (4) 2 × 10–15 2. The self ionization constant for pure formic acid , K = [ HCOOH + ] [HCOO ] has been estimated as 10 6 at room temperature .The density of formic acid is 1.15 g/cm3. The percentage of formic acid converted to formate ion are : (1) 0.002 % (2) 0.004 % (3) 0.006 % (4) 0.008 %  3. What is the Kb of a weak base that can produce one OH per molecule if its 0.04 M solution is 2.5% ionized. (1) 7  108 (2) 1.6  106 (3) 2.5  105 (4) 2  1011 4. [Cl¯ ] in a mixture of 200mL of 0.01 M HCl and 100 ml of 0.01 M BaCl2 is : (1) 0.01 M (2) 0.0133 M (3) 0.03 M (4) 0.02 M 5. 10–2 mole of NaOH was added to 10 litre of water. The pH will change by (1) 4 (2) 3 (3) 11 (4) 7 6. Blue litmus turns red in the following mixture of acid and base : (1) 100 mL of 1 × 10–2 M H SO + 100 mL of 1 × 10–2 M Ca (OH) (2) 100 mL of 1 × 10–2 M HCl + 100 mL of 1 × 10–2 M Ba (OH) (3) 100 mL of 1 × 10–2 M H SO + 10 mL of 1 × 10–2 M NaOH (4) 100 mL of 1 × 10–2 M HCl + 100 mL of 1 × 10–2 M NaOH 7. K for HCN is 5 x 10–10, calculate K for CN–. 8. K for trimethylamine is 6.4 x 10–5. Calculate K for trimethyl ammonium ion (CH ) NH+. 9. For the following equilibrium : NH + H O NH + + OH– equilibrium constant is 5.55 x 10–10. Calculate equilibrium constant for the equilibrium, NH + + H O NH OH + H+ 10. If equilibrium constant of CH COO– + H O CH COOH + OH– is 5.55 x 10–10, calculate equilibrium constant of CH COOH + H O CH COO– + H O+. 11. CO in aqueous solution shows following ionic equilibrium : 2H O + CO HCO – + H O+ If hydronium ion (H O+) concentration, is 2 x 10–6 M, what is hydroxide ion (OH–) concentration ? 12. Several acids are listed below with their respective equilibrium constants. HF(aq) + H O(𝑙) H O+ (aq) + F (aq) K = 7.2 x 10 4 HS (aq) + H O(𝑙) H O+ (aq) + S2  (aq) K = 1.3 x 10 11 CH COOH(aq) + H O(𝑙) H O+(aq) + CH COO (aq) K = 1.8 x10 5 3 2 3 3 a (i) Which is the strongest acid ? Which is the weakest ? (ii) What is the conjugate base of the acid HF ? (iii) Which acid has the weakest conjugate base ? (iv) Which acid has the strongest conjugate base ? Integer Answer Type This section contains 2 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. 13. In the following reaction, 3A + B 2C + D, initial moles of B is four times of A. At equilibrium moles of A and C are equal. Find % dissociation of B. 14. CH3–CO–CH3(g) CH3–CH3(g) + CO(g) 1 Initial pressure of CH3COCH3 is 12 atm. When equilibrium is set up mole fraction of CO(g) is 3 . Find Kp. ANSWER KEY DPP No.-43 1. 4 2. 1,3 3. 3 4. 4 5. 4 6. 2 7. 3 8. 1 9. 4 10. K = 1010. 11. 4 12. 1 13. 2 DPP No.-44 10. (i) 2NaOH + 2NO2  NaNO2 + NaNO3 + H2O ; 2NaOH + SO3  Na2SO4 + H2O. (ii) 6NaOH + 3Br2  5NaBr + NaBrO3 + 3H2O. ; 4NaOH + 2F2  4NaF + O2 + 2H2O. (iii) 6NaOH + 4S  2Na2S + Na2S2O3 + 3H2O. (iv) 2B + 6NaOH  2Na3BO3 + 3H2 (v) 2NaOH + Si + H2O  Na2SiO3 + 2H2 (vi) PbO + 2NaOH  Na2PbO2 + H2O ; PbO2 + NaOH  Na2PbO3 + H2O. (vii) 4NaOH + 2H2O + 2Al  2NaAlO2 + 3H2. (viii) Form insoluble hydroxides. CrCl3 + 3NaOH  Cr(OH)3  (Green) + 3NaCl. ; CuCl2 + 2NaOH  Cu(OH)2  (bule) + 2NaCl. (ix) HgCl2 + 2NaOH  Hg(OH)2  + 2NaCl ; Hg(OH)2  HgO  (yellow or brown) + H2O. 2AgNO3 + 2NaOH  2AgOH  + 2NaNO3 ; 2AgOH  Ag2O  (black) + H2O. 11. 4 12. 3 13. 2

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