106 small drops. The energy used will be (surface tension of mercury is 35 x10-3N/cm (a) 4.4 x 10-3 j (b) 2.2 x 10-4 j (c) 8.8 x 10-4 j (d) 104 j Ans. a (12)
103 Pa and 12 x 103 Pa, respectively. The containing 40 mole percent of A at this temperature is : a. xA = 0.28; xB = 0.72 b. xA = 0.76; xB = 0.24 c. xA = 0.37; xB = …
i. Pure solvent → separated solvent molecules ΔH1 ii. Pure solvent → separated solvent molecules ΔH2 iii. Separated solvent and solute molecules → solution ΔH3 Solution so formed will be ideal if a. ΔHsoln = ΔH3 – ΔH1 – ΔH2 b. ΔHsoln …
a. ΔG mix = 0 b. ΔH mix = 0 c. ΔU mix = 0 d. ΔP = P obs – P calculated by Raoult’s law = 0 Answer. a (14)
6.0 x 10 -4 atmosphere at 300K, the value of R used is 0.082 litre atmosphere K . The molecular -1 K -1 mass of polymer was found to be a. 3.0 x 102 b. 1.6 x 10 5 c. 5.6 x 104 d. …
10 ° C is 500 mm. Osmotic pressure of the solution become 105.3 mm. when it is dilute and temperauter raised to 25° C . the extent of dilution is a. 6 times b. 5 time c. 7 time d. …
KNO3 and CH3 COOH are prepared separately. Molarity of both is 0.1 M and osmotic pressures are P1 and P2 respectively. The correct relationship between the osmotic pressure is a. P2 > P1 b. P1 = P 2 c. P1 > …
dm3 of urea is isotonic with a 5 % solution of a nonvolatile solute. The molecular mass of this nonvolatile solute is a. 350 g mol-1 b. 200 g mol-1 c. 250 g mol-1 d. 300 g mol-1 Ans. d …
a. 200 mL of 2 M NaCl solution b. 200 mL of 1 M glucose solution c. 200 mL of 2 M urea solution d. all have osmotic pressure Ans. b (13)
a. 136. 2 b. 68.4 c. 34.2 d. 171. 2 Ans. b (5)
a. 0.1M gulcose b. o.1 M KCl c. 0.6% glucose solution d. 0.6 % KCl solution Ans. a Explanation : Isotonic solutions are those which have same concentration (13)
a. Equal temperature b. Equal osmotic pressure c. Equal volume d. Equal amount of solute Ans. b (5)
a. Sugar beet will lose water from its cells b. Sugar beet will absorb water from solution c. Sugar beet will neither absorb nor lose water d. Sugar beet will dissolve in solution Ans. a Explanation : Flow of liquid …
gcm-3 , molar mass of the substance will be a. 90.0 g mol -1 b. 115.0 g mol -1 c. 105.0 gmol -1 d. 210.0 gmol -1 Ans. d (8)
a. Isotonic b. Isomotic c. Hypertonic d. Equinormal Ans. a Explanation : Isotonic mixtures have equal osmotic pressure so there won’t be any flow of liquid between the blood cells and blood. (7)
15° C , then molecular weight of the solute is a. 1000 b. 1200 c. 1400 d. 1800 Ans. b (5)
37 ° C . What is the concentration of an aqueous NaCl a. 0.16 mol/L b. 0.32 mol / L c. 0.60 mol/ L d. 0. 45 mol/ L Ans. b (26)
a. Relative lowering of vapour pressure b. Elevation of boiling point c. Depression in freezing point d. Osmotic pressure e. Rast’s method Ans. d Explanation : Osmotic pressure method is especially suitable for the determination of molecular masses of macro …
a. A is less concentrated than B b. A is more same concentration c. Both have same concentratio d. None of these Ans. a Explanation : Osmosis occurs from dilute solution to concentrate solution. Therefore solution A is less concentrated …
a. Solute molecules only b. Solvent molecules only c. Solute and solvent molecules both d. Neither solute nor solvent molecules Ans. b (7)
(P1 ) , 10 g urea (P2 ) and 10g sucrose (P3) are dissolved in 250 mL of water is a. P1 >P2 > P3 b. P3 > P1 > P2 c.P2 >P1 > P3 d. P2 >P3 > P1 Ans. c (11)
(a) 100 : 1 (b) 1000 : 1 (c) 10 : 1 (d) 1 : 100 Ans. d (5)
The surface tension of liquid is 0.5 N/m. If a film is held on a ring of area 0.02 m2, its surface energy is (a) 5 × 10-2 joule (b) 2.0 × 10-2 joule (c) 4 × 10-4 joule (d) …
(a) Becomes zero (b) Becomes infinite (c) is equal to the value at room temperature (d) is half to the value at the room temperature Ans. a (9)
a. > 80 mL b. < 80 mL c. = 80 mL d. ≥ 80 mL Ans. b (23)
a. There will be no net movement across the membrane b. Glucose will flow across the membrane into urea solution c. Urea will flow across the membrane into glucose solution d. Water will flow from urea solution into glucose solution …
(a) 1000 : 1 (b) 1 : 1000 (c) 10 : 1 (d) 1 : 10 Ans. d (3)
a. Carbondisulphide – acetone b. Benzene – toluene c. Acetone – chloroform d. n- hexane – n- heptane e. Ethanol – acetone Ans. c Explanation : Acetone and chloroform form hydrogen bonds between he molecules of each other so in …
100° C) and HCl (b.p. 85° C) boils at 108.5 ° C. When this mixture is distilled it is possible to obtain a. Pure HCl b. Pure water c. Pure water as well as pure HCl d. Neither HCl nor …
a. Zero and zero b. +ve and zero c. -ve and zero d. -ve and -ve Ans. a (4)
a. Benzene + toluene b. n – hexane + n – heptane c. Ethyl bromide +ethyl iodide d. CCl4 + CHCl3 Ans. d (42)
(a) 10 cm per sec (b) 2.5 cm per sec (c) 5×4¹/3 per sec (d) 5×√2 per sec Answer-c (2)
a. CH3 COCH3 + CS2 b. C 6 H 6 + CH3 COCH3 c. CCl4+ CHCl3 d. CH3 COCH + CHCl3 Ans. d (7)
is 2 x 10-2 N/m. To blow a bubble of radius 1 cm, the work done is (a) 4 π x 10-6 J (b) 8 π x 10-6 J (c) 12 π x 10-6 J d) 16 π …
Mp . Whih of the following can be used to calculate molecular mass of the solute in terms of osmotic pressure (m = Mass of solute, V = Volume of solution and π = Osmotic pressure) a. Mp = (m/ π …
(a) R/2 (b) R/5 (c) R/6 (d) R/10 Ans. d (9)
(a) Rn(2/3) σ (b) (n(2/3) – 1) Rσ (c) (n(1/3) – 1) Rσ (d) 4πR2(n(1/3) – 1)σ (e) [1/( n(1/3) – 1)] σR Ans. d (4)
a. Heptane + Octane b. Water + Nitric acid c. Ethanol + water d. Acetone + Carbon disulphide Ans. b (11)