DPP-17 to 18 With Answer

REVISION QUESTIONS Single choice Objective ('–1' negative marking) Q.8 to Q.10 (3 marks 3 min.) [9, 9] 1. The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is (1) C > N > O > F (2) O > N > F > C (3*) O > F > N > C (4) F > O > N > C Sol. O (oxygen) have half filled outer orbitals. 2. The order of magnitude of ionic radii of ions Na+, Mg2+, Al3+ and Si4+ is (1) Na+ < Mg2+ < Al3+ < Si4+ (2) Mg2+ > Na+ > Al3+ > Si4+ (3) Al3+ > Na+ > Si4+ > Mg2+ (4*) Na+ > Mg2+ > Al3+ > Si4+ 3. Which of the following compound is isoelectronic with [NH  BH ] ? (1) B2H6 (2*) C2H6 (3) C2H4 (4) C3H6 Sol. (2) has same number of electrons i.e., 18. [NH  BH ] = 10 + 8 = 18. 4. What is the normality of the H2SO4 solution, 18.6 mL of which neutralizes 30.0 mL of a 1.55 N KOH solution? (1) 5.0 N (2) 1.25 N (3*) 2.5 N (4) 3.5 N 5. How many millilitres of a 9 N H2SO4 solution will be required to neutralize completely 20 mL of a 3.6 N NaOH solution? (1) 18.0 mL (2*) 8.0 mL (3) 16.0 mL (4) 80.0 mL 6. An excess of NaOH was added to 100 mL of a ferric chloride solution. This caused the precipitation of 1.425 g of Fe(OH)3. Calculate the normality of the ferric chloride solution (1) 0.20 N (2) 0.50 N (3) 0.25 N (4*) 0.40 N 7. As2O3 is oxidised to H3AsO4 by KMnO4 in acidic medium. Volume of 0.02M KMnO4 required for this purpose by 1mmol of As2O3 will be - (1) 10 mL (2) 20 mL (3*) 40 mL (4) 80 mL Revision Questions 8. 120 g of solution containing 40% by mass of NaCl are mixed with 200 g of a solution containing 15% by mass NaCl. Determine the mass percent of sodium chloride in the final solution. (1*) 24.4% (2) 78% (3) 48.8% (4) 19.68% Sol. mass of NaCl in st solution = 120 × 0.4 = 48 g mass of NaCl in nd solution = 200 × 0.15 = 30 g Total mass of NaCl = 30 + 48 = 78 g Total mass of solution = 120 + 200 = 320 g mass % of NaCl = 78  100 320 = 24.375 % 9. What is the molarity of solution if density of solution in 1.6 g/ml. (1) 5.5 M (2*) 6.6 M (3) 2.59 M (4) None 78 Sol. Molarity = 58.5 320 1.6 × 1000 = 6.66 M 10. Percentage (weight / vol) of NaCl present in the solution. (1) 24.4 % (2) 40% (3*) 39% (4) 3.9% Sol. % w/v = 78 × 100 = 39% 320 1.6 Subjective Questions ('–1' negative marking) Q.6 to Q.7 (4 marks 5 min.) [8, 10] REVISION QUESTIONS Single choice Objective ('–1' negative marking) Q.8 (3 marks 3 min.) [3, 3] Subjective Questions ('–1' negative marking) Q.9 (4 marks 5 min.) [4, 5] 1. What is the potential through which an electron, with a de Broglie wavelength of 1.5Å, should be accelerated, if its de Broglie wavelength would be reduced to 1 Å? (1) 110 volts (2) 70 volts (3*) 83 volts (4) 55 volts 2. In the isoelectronic species the ionic radii (Å) of N3–, O2– and F– are respectively given by : (1) 1.36, 1.40, 1.71 (2) 1.36, 1.71, 1.40 (3*) 1.71, 1.40, 1.36 (4) 1.71, 1.36, 1.40 Sol. On moving along the period ionic radii decreases. 3. Which has maximum first ionization potential. (1) C (2*) N (3) B (4) O Sol. Nitrogen have half filled p-orbital. 4. Which ion has greatest radius in the following : (1) H– (2) F– (3) Br– (4*) I– 5. Which one of the following is the smallest in size : (1) N3– (2) O2– (3) F – (4*) Na+ 6. Predict total spin for each configuration. (a) 1s2 (b) 1s2, 2s2 2p6 (c) 1s2, 2s2 2p5 (d) 1s2, 2s2 2p3 (e) 1s2 2s2 2p6, 3s2 3p6 3d5, 4s2. Ans. (a) 0 × (± 1/2) = 0 (b) 0 × (± 1/2) = 0 (c) 1 × (± 1/2) = ± 1/2 (d) 3 × (± 1/2) = ± 3/2 (e) 5 × (± 1/2) = ± 5/2 7. A neutral atom of an element has 2K, 8L, 9M and 2N electrons. Find out the following : (a) Atomic number (b) Total number of s electrons (c) Total number of p electrons (d) Total number of d electrons (e) Valency of element (f) Number of unpaired electrons Ans. (a) 21 (b) 8 (c) 12 (d) 1 (e) +2, +3 (f) 1. Sol. E.C.  1s2,2s2,2p6,3s2,3p6,3d1,4s2 Revision Questions 8. 50 mL of 5.6% KOH (w/v) is added to 50 mL of a 5.6% HCI (w/v) solution. The resulting solution will be (1) neutral (2) alkaline (3) strongly alkaline (4*) acidic 9. A sample of H2SO4 (density 1.787 g mL–1) is labelled as 86% by weight. What is molarity of acid ? What volume of acid has to be used to make 1 litre of 0.2 M H2SO4 ? Ans. 15.68 M, 12.75 m𝑙 86 1.787 Sol. M = 98 × 10 = 15.68 M M1V1 = M2V2 15.68 × V1 = 0.2 × 1000 V1 = 12.75 ml