271.Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of pure A and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is :- (1)xA = 0.28 ; xB = 0.72
- (2)xA = 0.76 ; xB = 0.24
- (3)xA = 0.37 ; xB = 0.63
- (4)xA = 0.4 ; xB = 0.6
272.For the solution of the gases w,x,y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is:- (1)
- (2)
- (3)
- (4)
273.What is the molarity of H2SO4 solution, that has a density 1.84 g/cc at 35oC and contains 98% by weight?- (1)18.4 M
- (2)18 M
- (3)4.18 M
- (4)8. 14M
274.For a dilute solution containing 2.5 g of a non- volatile non- electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol–1)- (1)724
- (2)740
- (3)736
- (4)718
275.Which one of the following is incorrect for ideal solution?- (1)∆Hmix=0
- (2)∆Umix=0
- (3)by Raoult’s law =0
- (4)∆Gmix=0
276.Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is:- (1)4A
- (2)2A
- (3)3A
- (4)A
277.Which one of the following electrolytes has the same value of Van’t Hoff factor (i) as that of Al2(SO4)3 (if all are 100% ionised)?- (1)Al(NO3)3
- (2)K4[Fe(CN)]
- (3)K2SO4
- (4)K3[Fe(CN)]
278.At 1000C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb=0.52, the boiling point ofthis solution will be- (1)102C
- (2)103C
- (3)101C
- (4)100C
279.Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing ΔTf, depression in the freezing point of a solvent in a solution?- (1)
- (2)
- (3)
- (4)
280.How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3.- (1)70.0 g conc. HNO3
- (2)54.0 g conc. HNO3
- (3)45.0 g conc. HNO3
- (4)90.0 g conc. HNO3
281.A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The apparent percentage dissociation of Na2SO4 is :- (1)25%
- (2)50%
- (3)75%
- (4)85%
282.A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is :(Molar mass of Cl = 35.5 g mol–1)- (1)0.675
- (2)0.162
- (3)0.486
- (4)0.325
283.Molarity of a 50 mL H2SO4 solution is 10.0 M. If the density of the solution is 1.4 g/cc, calculate its molality- (1)7.14
- (2)8.00
- (3)10.0
- (4)23.8
284.A solution contains non volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?- (1)M2=(πm2)VRT
- (2)M2=(Vm2)πRT
- (3)M2=(Vm2)πRT
- (4)M2=(Vm2)RTπ (m2= mass of solute, V= volume of solution, π= osmotic pressure)
285.Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressure of pure A and pure B are 7 × 10³ Pa and 12 × 10³ Pa, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is :- (1)xA=0.37;xB=0.63
- (2)xA=0.28; xB=0.72
- (3)xA=0.4;xB=0.6
- (4)xA=0.76;xB=0.24
286.A solution of acetone in ethanol- (1)obeys Raoult’s law
- (2)shows a negative deviation from Raoult’s law
- (3)shows a positive deviation from Raoult’s law
- (4)behaves like a near ideal solution.
287.Given below are two statements : one is labelled as Assertion (a) and the other is labelled as Reason (R) Assertion (a) : At 10°C, the density of a 5M solution of KC1 [atomic masses of K and Cl are 39 & 35.5 g mol−1. The solution is cooled to -21°C. The molality of the solution will remain unchanged. Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature. In the light of the above statements, choose the correct answer from the options given below:- (1)Both (a) and (R) are true and (R) is the correct explanation of (a)
- (2)Both (a) and (R) are true but (R) is not the correct explanation of (a)
- (3)(a) is true but (R) is false
- (4)(a) is false but (R) is true
288.A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute ' A ' of molar mass 60gmol−1 and 0.9 g of a non-volatile non-electrolyte solute ' B’ ' of molar mass 180gmol−1 in 100mLH2O at 27∘C. Osmotic pressure of the solution will be [Given: R=0.082LatmK−1mol−1]- (1)0.82 atm
- (2)2.46 atm
- (3)1.23 atm
- (4)1.47 atm
289.Choose the correct statement.- (1)KH value is same for a gas in any solution.
- (2)Higher the KH value more the solubility of gas.
- (3)KH value increases on increasing the temperature of the solution.
- (4)none of the above
290.Consider the following aqueous solutions. (I) 2.2 g Glucose in 125 mL of solution. (II) 1.9 g Calcium chloride in 250 mL of solution. (III) 9.0 g Urea in 500 mL of solution. (IV) 20.5 g Aluminium sulphate in 750 mL of solution. The correct increasing order of boiling point of these solutions will be : [Given : Molar mass in gmol−1:H=1,C=12,N=14,O=16,Cl=35.5,Ca=40,Al=27 and S=32 ]- (1)III <I<II< IV
- (2)I < II < III < IV
- (3)II < III < I < IV
- (4)II < III < IV < I
291.1.24 g of AX2 (molar mass 124gmol−1 ) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156∘C, while 25.4g∘ of AY2 (molar mass 250gmol−1 ) in 2 kg of water constitutes a solution with a boiling point of 100.0260∘C.Kb(H2O)=0.52Kkgmol−1 Which of the following is correct?- (1)AX2 is fully ionised while AY2 is completely unionised.
- (2)AX2 is completely unionised while AY2 is fully ionised.
- (3)AX2 and AY2 (both) are completely unionised.
- (4)AX2 and AY2 (both) are fully ionised.
292.’W ' g of a non-volatile electrolyte solid solute of molar mass ’M’gmol−1 when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and Kb for water is 0.52Kkgmol−1, then the mole fraction of the electrolyte solute (x2) in the solution can be expressed as (Given : density of water =1g/mL and boiling point of water =373K )- (1)162⋅6×WM
- (2)2⋅616×MW
- (3)81.3×WM
- (4)81.3×MW
293.For an ideal binary liquid mixture- (1)ΔS(mix)=0;ΔG(mix)=0
- (2)ΔH(mix)=0;ΔS(mix)<0
- (3)ΔV(mix)=0;ΔG(mix)>0
- (4)ΔS(mix)>0;ΔG(mix)<0
294.Match List-I and List-II. List-IA. OsmosisB. Reverse osmosisC. Electro osmosisD. ElectrophoresisList-III. Solvent molecules pass through semi permeable membrane towards solvent side.II. Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.III. Solvent molecules pass through semi permeable membrane towards solution sideIV. Dispersion medium moves in an electric field. Choose the correct answer from the options given below:- (1)A-I,B-III,C-IV,D-II
- (2)A-III, B-I, C-IV, D-II
- (3)A-III, B-I, C-II, D-IV
- (4)A-I, B-III, C-II, D-IV
295.The following solutions were prepared by dissolving 10 g of glucose (C6H12O6) in 250 ml of water (P1),10g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose (C12H22O11) in 250 ml of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is:- (1)P1>P2>P3
- (2)P2>P3>P1
- (3)P3>P1>P2
- (4)P2>P1>P3
296.Isotonic solutions are having the same- (1)surface tension
- (2)vapour pressure
- (3)osmotic pressure
- (4)viscosity:
297.Isotonic solutions have same- (1)Freezing temperature
- (2)Osmotic pressure
- (3)Boiling temperature
- (4)Vapour pressure
298.Evaluate the following statements for their correctness. (A) The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea. (B) Azeotropic mixtures boil without change in their composition (C) Osmosis always takes place from hypertonic to hypotonic solution (D)The density of 32% H2SO4 solution having molarity 4.09 M is approximately 1.26 g mL−1 (E) A negatively charged sol is obtained when KI solution is added to silver nitrate solution. Choose the correct answer from the options given below;- (1)B, D, and E only
- (2)A, B, and D only
- (3)A and C only
- (4)B and D only
299.A sample of water is found to contain 5.85%(w/W) of AB (molecular mass 58.5) and 9.50%(w/W)XY2 (molecular mass 95). Assuming 80% ionisation of AB and 60% ionisation of XY2, the freezing point of water sample is [Given: Kf for water is 1.86kkgmol−1. Freezing point of pure water is 273K and A,B and Y are monovalent ions.)- (1)264.25K
- (2)265.56K
- (3)280.44K
- (4)281.75K
300.Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2. When 1 g of PQ is dissolved in 50 g of solvent ' A’,ΔTb was 1.176 K while when 1 g of PQ2 is dissolved in 50 g of solvent ’A’,ΔTb was 0.689 K . ( Kb of ' A’=5Kkgmol−1 ). The molar masses of elements P and Q (in gmol−1 ) respectively, are:- (1)70,110
- (2)65,145
- (3)25,60
- (4)60,25
|