a. 1.5 M b. 1.66 M c. 0.017 M d. 1.59 M Ans. b (8)
H2 O2 solution of ’10 volume’ is a. 30% b. 3% c. 1% d. 10% Ans. b (6)
FeCl3 (formula mass = 162 ) has a density of 1.1 g/mL and contains 20.o% FeCl3 . Molar concentration of this solution is a. 0.028 b. 0.135 1.27 d. 1. 47 Ans. b (13)
a. x= y b. x = 2y c. x = 2y d. None of the above Ans. b (20)
a. 0.01 M b. 0.10 M c. 0.001 M d. 0.0001 M Ans. b Explanation : Molarity is intrinsic property. Hence it is independent of amount of solution (17)
HNO3 should be used to prepare 250mL of 2.0 M HNO3 a. 53. 0 g conc. HNO3 b. 45.0 g conc. HNO3 c.90.0 g conc. HNO3 d. 70.0 g conc. HNO3 Ans. b (12)
AgNO3 which of the following will be formula of the chloride (X stands for the symbol of the element other than chlorine) a. X2 Cl2 b. XCl2 c. XCl4 d. X 2Cl Ans. b (4)
H2 SO4 (Molar mass = 98 g mol -1 ) by mass will be a. 1.64 b. 1.88 c. 1.22 d. 1.45 Ans. c (81)
0.9 g L-1 , what will be the molarity of glucose in blood a. 5 M b. 50 M c. 0.005 M d. 0.5 M Ans. c (8)
a. 55.6 b. 5.56 c. 100 d. 20.08 Ans. a Explanation : molarity of pure water = 1000/18 = 55.6 M (6)
a. 15.25 b. 16.75 c. 18.92 d. 20.08 Ans. c (38)
H2 SO4 by mass and has a density of 1.80 g.mL-1 . Volume of acid required to make 1 litre of 0.1 M H2 SO4 solution. a. 11.10 mL b. 16.65 mL c. 22.20mL d. 5.55 mL Ans. d (44)
a. 12. 64 % b. 37. 92% c. 0.87% d. 63.21% Ans. b (7)
a. Add 0.585 g NaCl b. Add 20 mL water c. Add 0.010 mL NaCl d. Evaporate 10 mC water Ans. a (5)
a. 3.0504 b. 3.64 c. 3.o5 d. 2. 9732 Ans. d (6)
a. 0.07 b. 0. 14 c. 0. 28 d. 0.35 Ans. a (3)
M H2 SO4 a. 2 litre of the solution contains 0.020 mole of SO42- b. 2 litre of the solution contains 0.080 mole of H3 O+ c. 1 litre of the solution contains 0.020 moleH3 O+ d. None of these …
a. 0.2 b. 0.4 c. 1.0 d. 0.1 Ans.a (14)
a. 0.0099 b. 0.099 c. 0.99 d. 9.9 Ans. a (49)
NH3 is present in 250 mL volume. Its active mass is a. 1.0 mL-1 b. 0.5 mL-1 c. 1.5 mL-1 d. 2.0 mL-1 Ans. d (2)
a. 100 g b. 180 g c. 18 g d. 1.8 g Ans. a Explanation : 1og. of glucose is presnt in 100 mL of solution So 100 g of glucose is present in 1000 mL of solution (14)
a. 4:9 b. 2:3 c. 3:2 d. 1:1 Ans. b (8)
a. 1.78 M b. 2.00 M c. 2.05 M d. 2.22 M Ans. c (9)
a. 0.8 M b. 0.6 M c. 0.4 M d. 0.2 M Ans. b (8)
a. 900 mL b. 700 mL c. 500 mL d. 300 mL Ans. b (41)
H2 SO4 solution that has a density 1.84 g/cc at 35° C and contains solute 98% by weight a. 4.18 M b. 8.14 M c. 18.4 M d. 18 M Ans. c (13)
a. 480 g b. 48 g c. 38 g d. 380 g Ans. a (6)
10 20 molecules of urea are present in 100 mL of its solution. The concetration of urea solution is a. o.02 M b. 0.01 M c. 0.001 M d. 0.1 M Ans. b (3)
CH3 OH solution is very nearly equal to which of the following . a. 0.01% CH3 OH b.0.01m CH3 OH c .xCH3 OH = 0.01 d. 0.99 M H2 O e. 0.01 N CH3 OH ANSWER : E Explanation : For methyl alcohol N= M (16)
a. 18 litre b.9 litre c. 0.9 litre d. 1.8 litre Answer. D (8)
Na2 CO3 having 10.6g/500mL of solution is a. 0.2M b. 2M c.20M d. 0.02 M Ans. a (11)
a. Molarity b. Molality c. Mole fraction d. Weight fraction Ans. a (11)
AgNO3 what amount of AgNO3 should be added in 60 mL of solution a. 1.8 b. 0.8 c. 0.18 d. None of these Ans. a Explanation : Amont of AgNO3 added in 60 mL of solution = 60 x 0.03 …
a. 1.20 M b. 1.50M c. 1.344 M d. 2.70 M Ans. c (18)
H2 SO4 (36N) with 50mL of water is a. 36M b. 18M c. 9M d. 6M Ans. c Explanation : N1 V1 = N 2 V2, 36 x 50 = N 2 x 100 N2 = 36×50/100 = 18; 18 N H …
a. 0.5 b. 1.0 c. 2.0 d. 0.1 Ans. a (6)
Na2 CO3 present in 250 mL of 0.25 M solution is a. 6.225g b. 66.25 g c. 6.0 g d. 6.625g Ans. d (7)
The gas with the highest value of Hnery’s law constant is a. G4 b. G2 c. G3 d. G1 Ans. d (8)