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NEET Chemistry MCQ
Solutions And Liquid States Mcq Neet Chemistry
Quiz 5
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Q.1
The boiling point of 0.1 molal K4[ Fe(CN)6] solution will be ( Given Kb for water=0.52°C kg mol- ) .. [ Haryana CEET 1998]
0%
a) 100.52°C
0%
b) 100.104°C
0%
c) 100.26°C
0%
d) 102.6°C
Explanation
K4[ Fe(CN)6] ⇌ 4K+ + [ Fe(CN)6]4-, i=5 ΔTf=i Kf m =5×0.62×0.1=0.260°C Tb=100.26°CAnswer: (c)
Q.2
The molar freezing point constant for water is 1.86°C/mole. If 342g of cane sugar ( C12H22O11) is dissolved in 1000g of water, the solution will freeze at ... [ CPMT 1989]
0%
a) -7.44 °C
0%
b) -5.58°C
0%
c) -1.86°C
0%
d) -2.79°C
Explanation
342g of cane sugar dissolved in 1000g water=1 molal solution and i=1 ΔTf=i×Kfm =1×1.86 × 1=1.86 ∴ Tf=0 - 1.86=-1.86°CAnswer: (c)
Q.3
The vapour pressure of a pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and total vapour pressure of the solution is 84 torr at 27°C. The vapour pressure of pure liquid B at 27°C [ Pb.CET 1986]
0%
a) 14
0%
b) 56
0%
c) 140
0%
d) 70
Explanation
Answer: (c)
Q.4
The concentration of an aqueous solution of 0.01 M CH3OH solution is very nearly equal to which of the following? [ BIT 1992]
0%
a) 0.01% CH3OH
0%
b) 0.01 m CH3OH
0%
c) mole fraction of CH3OH 0.01
0%
d) 0.99 M H2O
Explanation
Due to low concentration, 0.01 M CH3OH ≡ 0.01 m CH3OH Answer: (b)
Q.5
What is the freezing point of the solution containing 8.1g of HBr in 100 g water assuming the acid to be 90% ionised ( Kf for water=1.86 K kh mol-) .. [ BHU 1981]
0%
a) 0.85°C
0%
b) -3.53°C
0%
c) 0°C
0%
d) -0.35°C
Explanation
On dissociation of HBr it gives two ions Since dissociation is 90% there fore 0.1 HBr remains undissociated and have produced 0.9 ions of H+ and 0.9 ions of Br-Thus to total particles=0.1+ 0.9+ 0.9=1.9or i=1.9 Now 8.1 g of HBr=8.1 /81=0.1, wt of solvent=0.1kg Thus molality m=0.1/0.1=1 ΔTf=i Kf m =1.9 ×1.86=3.53°Freezing point of pure water is=0° Thus freezing point of solution=0-3.53=-3.53°CAnswer: (c)
Q.6
4.0 of NaOH is dissolved in 100cc solution. The normality os solution is .. [ MP PMT 1995]
0%
a) 1.0
0%
b) 0.1
0%
c) 0.5
0%
d) 0.4
Explanation
Acidity of NaOH=1 ∴ equ. wt=40 Answer: (a)
Q.7
The number of atoms of iodine atoms present in 1cm3 of its 0.1M solution is ... [ PB.CET 1990]
0%
a) 3.02 ×1023
0%
b) 6.02×22
0%
c) 6.02×1019
0%
d) 1.204×1020
Explanation
0.1M solution or 0.1 mole in one litre Thus in 1cc or 1ml will have 0.1×103 or 10-4 moles of Iodine (I2)∴ number of atoms=2× 10-4 ×6.02×23Answer: (d)
Q.8
The depression in freezing point of 0.01M aqueous solutions urea, sodium chloride and sodium sulphate is in the ratio of ... [ DEC 1994]
0%
a) 1 : 1 : 1
0%
b) 1 : 2 : 3
0%
c) 1 : 2 : 4
0%
d) 2 : 2 : 3
Explanation
Concentration of particles in 0.01M solution of urea, sodium chloride and sodium sulphate is 0.01M, 0.02M and 0.03M respectively They are in ratio 1:2:3. Hence depression in freezing point will be sameAnswer: (b)
Q.9
A molal solution of a sodium chloride has a density of 1.21 g/ml. The molarity of this solution is ... [ DEC 1994]
0%
a) 4.15
0%
b) 1.143
0%
c) 2.95
0%
d) 3.15
Explanation
1 m NaCl solution=1 mole NaCl ( 58.5g) in 1000g water Thus wt of 1m solution=58.5 +1000=1085.5 g solutiondensity given 1.21 ∴ volume=1058.5/121=875 mlThus one mole NaCl in 875 mlNow molarity of solution=(1/875)×1000=1.143Answer: (b)
Q.10
The molal b.p. constant for water is 0.513°C Kg mol-. When 0.1 mole of sugar is dissolved in 200g of water, the solution boils under a pressure of 1atm at ... [ AIIMS 1991]
0%
a) 100.513°C
0%
b) 100.0513°C
0%
c) 100.256°C
0%
d) 10.1.025°C
Explanation
0.1 mole of sugar is dissolved in 200g of water=0.5 molal solution °Tb=Kb × m=0.513 × 0.5=0.2565°CTb=100 +0.2565=100.2565°cAnswer: (c)
Q.11
The normality of 10% ( weight/volume) acetic acid is .... [ CPMT 1983 ]
0%
a) 1 N
0%
b) 10 N
0%
c) 1.7 N
0%
d) 0.83 N
Explanation
10% wr/vol means 10g in 100cc Eq weight of CH3COOH = 60 Answer: (c)
Q.12
The vapour pressure of water at room temperature is 23.8 mmHg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to .. [ Pb CET 1986 ]
0%
a) 23.9 mmHg
0%
b) 24.2 mmHg
0%
c) 21.42 mmHg
0%
d) 21.44 mmHg
Explanation
Answer: (c)
Q.13
An aqueous solution of sucrose. C12H12O11, containing 34.2 g/l has an osmotic pressure of 2.38 atm at 17°C. For an aqueous solution of glucose, C6H12O6 to be isotonic with this solution, it would have ... [ AMU 1997 ]
0%
a) 34.2 g/l
0%
b) 17.1 g/l
0%
c) 18.0 g/l
0%
d) 36.0 g/l
Explanation
34.2 g/l sucrose = 0.1 M 18g/l glucose = 0.1M Answer: (c)
Q.14
10ml of concentrated sulphuric scid ( 18 molar) is diluted to 10 litre. The approximate strength of acid would be ...[ JIPMER 1994 ]
0%
a) 0.18 N
0%
b) 0.036 N
0%
c) 0.36N
0%
d) 0.09 N
Explanation
18 molar = 36 N sulphuric acid N1V1 = N2V2 36 × 10 = N2 × 10,000 N2 = 0.036N Answer: (b)
Q.15
A solution containing 3.3g of a substance in 125g of benzene ( b.p 80°C) boils at 80.66°C. If Kb for one litre of benzene is 3.28°C, the molecular weight of the substance shall be ... [ CPMT 1992 ]
0%
a) 127.20
0%
b) 131.20
0%
c) 137.12
0%
d) 142.72
Explanation
Answer: (b)
Q.16
The molal elevation constant of water = 0.52°C. The boiling point of 1.0 molal aqueous KCl solution ( assume complete dissociation of KCl) should be... [ BHU 1987 ]
0%
a) 100.52°C
0%
b) 101.04°C
0%
c) 99.48°C
0%
d) 98.96°C
Explanation
ΔTb = i Kb m = 2×0.52×1= 1.04° Boiling point of pure water = 100°C Thus Boiling point of solution = 100 + 1.04 = 101.04°C Answer: (b)
Q.17
What is the molality of solution of a certain solute in a solvent if their is a freezing point depression of 0.184°C and if the freezing point constant is 18.4... [ Haryana CEET 1991 ]
0%
a) 0.01
0%
b) 1
0%
c) 0.001
0%
d) 100
Explanation
i = 1 ΔTf = i×Kfm 0.184 = 1×18.4 ×m m = 0.01 Answer: (a)
Q.18
An aqueous solution containing 1g of urea boils at 100.25°C. The aqueous solution containing 3g of glucose in the same volume will boil at ... [ BHU 1994 ]
0%
a) 100.75°C
0%
b) 100.5°C
0%
c) 100.0°C
0%
d) 100.25°C
Explanation
1g urea = (1/60) mol, 3g glucose = 2/180 = (1/60) mole Hence it will boil at the same temperature Answer: (d)
Q.19
An aqueous solution freezes at -0.186°C ( Kf = 1.86, Kb = 0.512 ). What is the elevation in boiling point? [ AIEEE 2002 ]
0%
a) 0.186
0%
b) 0.512
0%
c) 0.86
0%
d) 0.0512
Explanation
0.186 = 1.86 × m or m = 0.1 ΔTb = 0.512×0.1 = 0.0512°C Answer: (d)
Q.20
A 0.001 molal solution of [Pt(NH3)4Cl4] in water had a freezing point depression of 0.0054°C. If Kf for water is 1.80, the correct formulation of the above molecule is .. [ kerala MEE 2003]
0%
a) [Pt(NH3)4Cl3]Cl
0%
b) [Pt(NH3)4Cl2]Cl2
0%
c)[Pt(NH3)4Cl]Cl3
0%
d) [Pt(NH3)4Cl4]
Explanation
[Pt(NH3)4Cl4] on dissociation produces n moles of particles ∴ i=n For the formula ΔTf=i Kf m0.0054=n ×1.8 × 0.001n=3Hence the formula must be the one which gives 3 ions of productAnswer: (b)
Q.21
The molality of sulphuric acid solution in which the mole fraction of water is 0.85 is ... [ Haryana CEET 1994]
0%
a) 9.80
0%
b) 10.50
0%
c) 10.58
0%
d) 11.25
Explanation
let mole fraction of water x1 be 0.85 and n1 are the number of moles and mole fraction of sulphuric acid=0.15 and n2 are number of molestaking 1000 of water then n1=1000/18 moleson substituting value of n1 in above equation we getn2=9.8Answer: (a)
Q.22
The molal freezing point constant for water is 1.86°C/m. therefore, the freezing point of 0.1M NaCl solution in water is expected to be ... [ MLNR 1994]
0%
a) -1.86°C
0%
b) -0.186°C
0%
c) -0.372°C
0%
d) +0.372°C
Explanation
ΔTf=i Kf m =2×1.86×0.1=0.372 Freezing point of water is 0°C∴ Tf=0 - 0.372=-0.372 °CAnswer: (c)
Q.23
The van't Hoff factor for 0.1M Ba(NO3)2 solution is 2.The degree of dissociation is ... [ IIT 1999]
0%
a) 91.3%
0%
b) 87%
0%
c) 100%
0%
d) 74%
Explanation
Ba(NO3)2 ⇌ Ba2+ + 2NO3- Thus 1 mole of Ba(NO3)2 gives 3 mole of ionsIf α moles of Ba(NO3)2 dissociated will produce ,α moles of Ba2+ and 2α moles of NO3-At equilibrium concentration will be (1-α), α, 2αor particle concentration at equilibrium will be 1+2α∴ i=1+ 2α2.74=1 + 2α or α=0.87=87%Answer: (b)
Q.24
A solution of 4.5g of pure non-electrolyte in 100g of water was found to freeze at -0.465°C. The molecular weight of the solute is closest to ( Kf=1.86 ) .. [ kerala CET 2003]
0%
a) 135.0
0%
b) 172.0
0%
c) 90.0
0%
d) 180.0
Explanation
Answer: (d)
Q.25
The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate will be ... [ BHU 1982]
0%
a) 60%
0%
b) 83.5%
0%
c) 46.7%
0%
d) 60.23%
Explanation
i=calculated molecular weights / observed molecular weight i=170/92.64=1.835α=i - 1=1.835 -1=0.835=83.5%Answer: (b)
Q.26
The osmotic pressure of 0.2 molar solution of urea at 27°C is ( R=0.082 litre atm mol-1 K-1) is ... [ Pb. CET 1986]
0%
a) 4.92 atm
0%
b) 1.0 atm
0%
c) 0.2 atm
0%
d) 27 atm
Explanation
π=CRT=0.2×0.082×300=4.92 atm Answer: (a)
Q.27
If the elevation in boiling point of a solution of 10g of solute ( mol.wt.=100) in 100g of water is ΔTb, the ebulliscopic constant of water is ...
0%
a) 10
0%
b) 100Tb
0%
c) ΔTb
0%
d) ΔTb/10
Explanation
100gm solute=0.1mole in 100 gm water=1 molal solution ΔTb=Kb ×mΔTb=Kb ×1=KbAnswer: (c)
Q.28
The osmotic pressure of 5% ( mass-volume) solution of cane sugar at 150°C ( mol. mass sugar=342) is : .. [ BHU 1995]
0%
a) 4 atm
0%
b) 5.07 atm
0%
c) 3.55 atm
0%
d) 2.45 atm
Explanation
Concentration of cane sugar is 5% or 50 gram in litre or ( 50/342) in litre or 50/342 M Now Π=CRT=(50/342) × 0.082 × 423=5.07 atmAnswer: (b)
Q.29
Ina 0.2 molal aqueous solution of a weak acid HX, the degree of ionization is 0.Taking Kf for water as 1.85, the freezing point of the solution will be nearest to ... [ AIEEE 2003]
0%
a) -0.360°C
0%
b) -0.260°C
0%
c) +0.480°C
0%
d) -0.480°C
Explanation
HX ⇌ H+ + X+After dissociation concentration of HX=1 -0.3 Concentration of H+ and X- is 0.3 eachThus total concentration of particle=0.7+0.3+.03=1.3 or i=1.3 ΔTf=i Kf m=1.3×1.85×0.2=0.481°Tf=0 -0.481=-0.481°CAnswer: (d)
Q.30
The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is .. [ Pb. CET 1988]
0%
a) 0.50
0%
b) 0.60
0%
c) 0.27
0%
d) 0.73
Explanation
For equimolar solution xB=xT=0.5 pB=xB × pBo=0.5×160=80 mmpT=xT × pBo=0.5×60=30 mmTotal=80 + 30=110 mm Mole fraction of toluene in vapour phase=30/110=0.27Answer: (c)
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