1.The freezing point of a diluted milk sample is found to be −0.2°C, while it should have been −0.5°Cfor pure milk. How much water has been added to pure milk to make the diluted sample?- (1)1 cup of water to 2 cups of pure milk
- (2)3 cups of water to 2 cups of pure milk
- (3)1 cup of water to 3 cups of pure milk
- (4)2 cups of water to 3 cups of pure milk
2.Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how may grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8ºC?- (1)72 g
- (2)93 g
- (3)39 g
- (4)27 g
3.If aqueous solution contains 9% and 1%(w/w) of two non-volatile non-electrolytes X (molecular weight 180) and Y (molecular weight 50) respectively, the boiling point of solution in ∘C approximately is (Kb=0.52Kkgmol−1)- (1)101.4
- (2)100.4
- (3)102.4
- (4)100.8
4.A 6% solution of urea is isotonic with- (1)0.05M solution of glucose
- (2)6% solution of glucose
- (3)25% solution of glucose
- (4)1M solution of glucose.
5.Consider a solution of CO2(g) dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of CO2 in vapour phase above water) [y-axis] with log (mole fraction of CO2 in water) [x-axis ] at 25∘C ?- (1)
- (2)
- (3)
- (4)
6.The Molarity (M) of an aqueous solution containing 5.85g of NaCl in 500mL water is : (Given : Molar Mass Na:23 and Cl:35.5gmol−1)- (1)2
- (2)20
- (3)4
- (4)0.2
7.At T(K)×g of a non-volatile solid (molar mass 78g mol−1 ) when added to 0.5kg water, lowered its freezing point by 1.0∘C. What is x (in (g) ? (Kf) of water at T(K)= (1.86KKgmol−1)- (1)10.48
- (2)20.96
- (3)41.92
- (4)5.24
8.200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57×10−3 bar. The molar mass of protein will be(R=0.083L bar mol−1K−1)- (1)51022 gmol−1
- (2)122044 gmol−1
- (3)31011 gmol−1
- (4)61038 gmol−1
9.Which of the following pair of solutions is isotonic? A. 18g/L of glucose of solution and 6g/L of Urea solution B. 10g/L of glucose solution and 10g/L of Urea solution C. 0.01MNaOH solution and 0.02M glucose solution D. 0.01MNaCl solution and 0.01M glucose solution (Assume that NaCl undergoes complete dissociation)- (1)A and B
- (2)A and C
- (3)B and D
- (4)B and C
10.A solution of two liquids boils at a temperature more than the boiling point of either of them. Hence, the binary solution shows- (1)negative deviation from Raoult's law
- (2)positive deviation from Raoult's law
- (3)no deviation from Raoult's law
- (4)positive or negative deviation from Raoult's law depending upon the composition.
11.Assertion (A) : For an endothermic dissolution process, increase in temperature increases the solubility in a nearly saturated solution. Reason (R) : In saturated solution dynamic equilibrium exists between dissolved solute and undissolved solute.- (1)Both (A) and (R) are correct and R is the correct explanation of A
- (2)Both (A) and (R) are correct, but R is not the correct explanation of A
- (3)(A) is correct but ( R ) is not correct
- (4)(A) is incorrect but (R) is correct
12.If the KH values for Ar(g),CO2(g),HCHO(g) and CH4(g) respectively are 40.39,1.67,1.83×10−5 and 0.413 , then identify the correct increasing order of their solubilities.- (1)HCHO<CH4<CO2<Ar
- (2)HCHO<CO2<CH4<Ar
- (3)Ar<CO2<HCHO<CH4
- (4)Ar<CO2<CH4<HCHO
13.A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:- (1)1400mmHg,A
- (2)1400mmHg,B
- (3)600mmHg,A
- (4)600mmHg,B
14.At 298K, if the vapour pressure of pure liquids toluene, benzene, chloroform and dichloromethane are 60,160 , 200 and 415 torr respectively. Then which liquid is having high boiling point?- (1)Toluene
- (2)Benzene
- (3)Chloroform
- (4)Dichloromethane
15.Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2°C. The percentage association of solute A in water, is [Given : Molar mass of A = 93 g mol−1. Molal depression constant of water is 1.86 K kg mol−1]- (1)50 %
- (2)60 %
- (3)70 %
- (4)80 %
16.8 g of NaOH is dissolved in 18 g of H2O. Mole fraction of NaOH in solution and molality (in molKg−1) of the solution respectively are :- (1)0.2, 22.20
- (2)0.2, 11.11
- (3)0.167, 11.11
- (4)0.167, 22.20
17.0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mo1−1, the lowering in freezing point of the solution is- (1)0.56 K
- (2)1.12 K
- (3)−0.56K
- (4)−1.12K
18.What is the boiling point of solution of 0.1mKCl ? Kb of water is 0.52Kkgmol−1.(α=100%) (water boil at 373K)- (1)100.104K
- (2)373.104K
- (3)273.104K
- (4)373.052K
19.The mixture which shows positive deviation from Raoult's law is:- (1)Benzene + Toluene
- (2)Acetone + Chloroform
- (3)Chloroethane + Bromoethane
- (4)Ethanol + Acetone
20.Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700 mmHg, respectively, at the same temperature. Then correct statements is : (xM = Mole fraction of ‘M’ in solution ; xN = Mole fraction of ‘N’ in solution; yM = Mole fraction of ‘M’ in vapour phase; yN = Mole fraction of ‘N’ in vapour phase)- (1)xNxM=yNyM
- (2)(xM−yM)<(xN−yN)
- (3)xNxM<yNyM
- (4)xNxM>yNyM
21.When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is:- (1)0.5
- (2)1
- (3)2
- (4)3
22.The vapour pressure of a solvent decreased by 10 mm ofmercury when a non‐volatile solute was added to the solvent. The mole fraction ofthe solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?- (1)0.4
- (2)0.6
- (3)0.8
- (4)0.2
23.15% aqueous solution of glucose (molecular weight =180g/mol ) is isotonic with 8% aqueous solution containing an unknown non-dissociable solute. What is the molecular weight of the unknown solute?- (1)108
- (2)96
- (3)84
- (4)9.6
24.A solution containing components A and B follows Raoult’s law- (1)A‐B attraction force is greater than A‐A and B‐B
- (2)A‐B attraction force is less than A‐A and B‐B
- (3)A‐B attraction force remains same as A‐A and B‐B
- (4)volume of solution is different from sum of volume of solute and solvent.
25.The vapour pressure of water at 20ºC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2g of water at 20°C, the vapour pressure of the resulting solution will be- (1)15.750 mm Hg
- (2)16.500 mm Hg
- (3)17.325 mm Hg
- (4)17.675 mm Hg
26.The quantity which changes with temperature is:- (1)Molarity
- (2)Mass percentage
- (3)Molality
- (4)Mole fraction
27.At 298 K, the mole percentage of N2(g) in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N2(g) in water at 298 K? ( KH for N2 is 6.5×107mmHg )- (1)1.23×10−7
- (2)1.17×10−4
- (3)9.35×105
- (4)9.35×10−5
28.Solute ' X ' dimerises in water to the extent of 80%,2.5g of ' X ' in 100g of water increases the boiling point by 0.3∘C. The molar mass of ' X ' is (Kb=0.52Kkgmol−1)- (1)52
- (2)65
- (3)26
- (4)13
29.18 g glucosse (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is.- (1)76.0
- (2)752.4
- (3)759.0
- (4)7.6
30.34.2g of sugar was dissolved in water to produce 214.2g of sugar syrup. The molality and mole fraction of sugar in the syrup are respectively ( C=12,H=1,O=16 )- (1)0.224m,0.002
- (2)0.05m,0.02
- (3)0.556m,0.0099
- (4)0.05m,0.99
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