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NEET Chemistry MCQ
Solutions And Liquid States Mcq Neet Chemistry
Quiz 1
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Q.1
For a dilute solution, Roult's law states that .. [ IIT 1985]
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a) The lowering of vapour pressure is equal to mole fraction of the solute
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b) The relative lowering of vapour pressure is equal to the mole fraction of the solute
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c) The relative lowering of vapour pressure is equal to the amount of the solute in the solution.
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d) The vapour pressure of the solution is equal to mole fraction of the solvent.
Explanation
Answer: (b)
Q.2
Which of the following is colligative property .. [ AFMC 1992]
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a) Boiling point
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b) Vapour pressure
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c) Osmotic pressure
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d) Freezing point
Explanation
Answer: (c)
Q.3
Which of the following statement is correct? .... [ Pb.CET 1986]
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a) Lowering of vapour pressure takes place only in ideal solution
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b) Lowering of vapour pressure does not depend upon the solvent at a given concentration of the solute
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c) Lowering of vapour pressure depends upon the nature of the solute
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d) relative lowering of vapour pressure does not depend upon the solvent at a given concentration of the solute
Explanation
Answer: (d)
Q.4
Maximum freezing point falls isn ... [ MP PMT 1986]
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a) Camphor
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b) Napthalene
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c) Benzene
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d) Water
Explanation
Camphor has the maximum value of Kf( approx 40°) Answer: (a)
Q.5
mixture which shows positive deviation from Raoult’s law is .... [NEET 2020]
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a) Acetone + Chloroform
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b) Chloroethane + Bromoethane
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c) Ethanol + Acetone
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d) Benzene + Toluene
Explanation
Answer: (c)
Q.6
The freezing point depression constant (Kf) of benzene is 5.12 K kg mol-The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) : [NEET 2020]
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a) 0.40 K
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b) 0.60 K
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c) 0.20 K
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d) 0.80 K
Explanation
Answer: (a)
Q.7
The mixture that forms maximum boiling azeotrope is: [NEET 2019]
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a) Water + Nitric acid
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b) Ethanol + Water
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c) Acetone + Carbon disulphide
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d) Heptane + Octane
Explanation
A solution that shows large negative deviation from Raoult's law forms a maximum boiling azeotrope at a specific composition. Nitric acid and water is an example of this class of azeotrope. This azeotrope has an approximate composition of 68% nitric acid and 32% water by mass, with a boiling point of 393.5 K (120.4 °C). Answer: (a)
Q.8
One mole of a substance is present in 1 kg of solvent. The correct statement regarding above solution is [AFMC 1998]
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a) Strength gm/gm
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b) It shows molar concentration
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c) Molal concentration
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d) Normality
Explanation
Molality is Number of moles per kg of solvent Answer: (c)
Q.9
5 gm of CH3COOH is dissolved in one litre of ethanol. Suppose there is no reaction between them. If the density of ethanol is 0.789 gm/ml then the molality of resulting solution is [AFMC 1998]
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a) 0.0256
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b) 0.1056
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c)0.1288
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d)0.1476
Explanation
Here solvent is Ethanol Density of ethanol 0.789 gm/mlWeight of 1000ml Solvent=Volume×density=789 gmMolecular weight of CH3COOH=60 gmFormula for molality m : (w×1000)/(M0×W)Here Molecular weight of Solute M0=60gm Weight of solute w=5g Weight of Solvent W=789gBy substituting above values in equation (1) we gate Molality m=0.1056 mol/g Answer: (b)
Q.10
When 800 gm of a 40% solution by weight was cooled, 100 gm of solute precipitated. The percentage composition of remaining solutionis [AFMC 1998]
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a) 31.43%
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b) 5.56%
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c)6.78%
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d)6.96%
Explanation
Weight of solute present in solution before cooling =40% of 800=40/100 × 800=320 g100 gm is precipitated remaining wt of solute in solution=220gNow new weight of Solution=800-100=700g% by wt=(weight of solute)/(weight of solution) ×100=220/700 × 100=31.43% Answer: (a)
Q.11
Which of the following mode of expressing concentration is independent of temperature? [ IIT 1988]
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a) Molarity
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b) Molality
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c) Formality
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d) Normality
Explanation
Answer: (b)
Q.12
0.45 g of acid of molecular weight 90 was neutralized by 20 ml. of a 0.5N caustic soda. The basicity of an acid is [AFMC 1998] Solution:
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a)1
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b)2
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c)3
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d)4
Explanation
Here 0.45 g of acid of molecular weight 90 Hence Number of moles of acid Neutralized=5×10-2=0.005Equivalent weight of NaOH=40There fore weight of NaOH in 1 liter=40×0.5=20 gmWeight of NaOH in 0.020 liter=20×0.02=0.4g0.4g of NaOH=0.01 mole From above it is clear that 0.005 moles are neutralized by 0.01 moles of NaOH Number of moles of Base × Acidity=Number of moles of acid × basicity Therefore basicity of acid=2 Answer : (b)
Q.13
The freezing point of a 3% aq. solution of A is equal to the freezing point of 9% aq. solution of B. If the molecular weight of 'A' is 60, then the Molecular wt. of B is [AFMC 1998]
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a) )45
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b) 90
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c) 180
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d) 360
Explanation
Since freezing point is same for both solution hence ∆Tf is sameHence we getLet W1=W2=100gSince solvent is same Kf1=Kf2Now 3% solution of A w1=3gMolecular wt of A=Mo1=60Now 9% solution of B=w2=9gOn substituting values in above equation we get3/60=9/Mo2Mo2=180 Answer: (c)
Q.14
The semi-permeable membrane allows the passage of [AFMC 1998]
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a)Solvent particle
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b)Solute particle
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c)Solution
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d)All of these
Explanation
A: Semi-permeable membrane allows passage of solvent molecules Answer: (a)
Q.15
Tyndall effect can be observed in [AFMC 1999]
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a) colloidal solution
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b) solvent
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c)soluted
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d)precipitate
Explanation
Tyndall effect is scattering of light due to colloidal particles. Because of which path of light is visible. Answer: (a)
Q.16
The amount of dibasic acid present in 100 ml of the aq. solution to give solution of normality of 0.1 N is….[mol. wt.=200] [AFMC 1999]
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a)0.5 gm
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b)1 gm
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c)1.5 gm
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d)2 gm
Explanation
Normality=(w×Acidity or basicity×1000)/(Mo×V)Wt of solute w=?Molecular wt Mo=200 Volume=100ml=0.1 liter Basicity=20.1=(w ×2×1000 )/(200×0.1)On simplification we getW=1/1000=1 g Answer : (b)
Q.17
Isotonic solutions have [AFMC 1999]
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a) same vapour pressure
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b) same osmotic pressure
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c) same freezing point
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d) same temperature
Explanation
Solution having same osmotic pressure are called isotonic solutions. When two solutions of different solute concentrations are separated by a semipermeable membrane, the solvent molecules will move from high solute concentration to lower solute concentration to balance the osmotic pressure on both sides Answer: (b)
Q.18
Which is the colligative property? [AFMC 2000]
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a)Surface tension
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b)Osmotic pressure
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c)Viscosity
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d)Refractive index
Explanation
Osmotic pressure is a colligative property as it depend on the number of particles and not on the nature of solute.Answer: (b)
Q.19
Normality of 0.3 M H3 PO4 solution is [AFMC 2000]
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a) 0.3N
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b) 0.4N
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c)0.6N
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d)0.9N
Explanation
Normality=Molarity × Basicity=0.3×3=0.9NAnswer: (d)
Q.20
The solutions which have same osmotic concentrations are known as [AFMC 2001]
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a) Normal
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b) Isotonic
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c)Hypotonic
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d)Hypertonic
Explanation
Two Solutions having same osmotic pressure are called isotonic solutionOut of two solutions, solution having higher osmotic pressure is called Hypertonic And lower osmotic pressure is called Hypotonic solutions Answer :(b)
Q.21
Molality does not change with [AFMC 2001]
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a) Temperature
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b) Concentration
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c) Pressure
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d) All of these
Explanation
Molecular weight of Solute Mo Weight of solute=w Weight of Solvent=W All the quantities are mass hence do not depend on the temperatureAnswer: (a)
Q.22
4 gm caustic soda is dissolved in 100 cc of solution. The normality of solution is [AFMC 2001]
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a)0
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b)0.5
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c)1
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d)1.5
Explanation
A: Volume V=100cc=100 mlWeight of solute w=4 gramacidity of NaOH=1Molecular weight of NaOH M0=40 gramNormality N=(w×Acidity or basicity×1000)/(M0×V)By substituting values in above equation Normality N=1Answer: (c)
Q.23
10 gm of glucose is dissolved in 150 gm. of water.The mass percentage of glucose is [AFMC 2001]
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a) 2.50%
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b) 6.25%
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c)8.75%
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d)10%
Explanation
Mass % of glucose=(Mass of Glucose)/(Total mass)× 100Mass % of glucose=[10/150] × 100=6.25%Answer: (b)
Q.24
The molarity of mixture obtained by mixing 100 ml of 0.4 M H2SO4 and 200 ml of 0.2M HCl is ..[AFMC 2002]
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a)0.0267
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b)0.2670
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c)1.0267
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d)1.1670
Explanation
Molarity of mixture=(M1V1 +M2V2)/ (V1 + V2) =(40 +40)/300 =80/300=0.267 M Answer : (b)
Q.25
Buffer solutions can be obtained by mixing aqueous solution of [AFMC 2002, 2004]
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a) NaOH and HCl
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b) CH3COOH and NaOH
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c) CH3COONa and CH3COOH
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d) CH3COONa and HCl
Explanation
Buffer solution resists change in pHIn case of CH3COONa and CH3COOH If we add H+ then it will combine CH3COO- of CH3COONa to given CH3COOHIf we add OH- the it will react with Na+ to give NaOH Answer : (c)
Q.26
Brownian movement is found in [AFMC 2002]
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a)unsaturated solution
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b) saturated solution
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c)colloidal solution
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d)suspension solution
Explanation
Random motion of colloidal particle is called as Brownian motion. Colloidal particle follows zig-zag path in liquid medium due to collision of liquid molecules with colloidal particles Answer: (C)
Q.27
A solution is formed by diluting 250 ml. of 0.40 N H2SO4 with one litre of water. The normality of above formed solution is : [AFMC 2003]
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a)0.4N
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b) 0.899N
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c)0.04N
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d)0.08N
Explanation
N1V1=N2V2 Here N1=0.4 ; V1=250ml ; V2=250+1000 ml, water added After substituting values in equation we get N2=0.08N Answer : (d)
Q.28
The term used for diffusion of solvent through a semi-permeable membrane is known as: [AFMC 2003]
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a) osmosis
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b) plasmolysis
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c)diffusion
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d)active absorption
Explanation
Flow of solvent through a semipermeable membrane from solution with low concentration of solute or pure solvent to Higher concentration solute solution Answer: (a)
Q.29
Distribution law was given by [AFMC 2003]
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a) Ostwald
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b) van't Hoff
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c) Nernst
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d) Henry
Explanation
Nernst distribution law:At constant temperature when different quantities of a solute are allowed to distribute between two immiscible solvents in contact with each other, then at equilibrium the ratio of concentration of the solute in two layers is constant CA / CB=K0=Partition or distribution co-efficient Answer : (c)
Q.30
he molar concentration of 20g of NaOH present in 5 litre of solution is [AIIMS 1998]
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a)0.1moles/litre
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b)0.2moles/litre
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c)1 moles/litre
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d)2 moles/litre
Explanation
molar concentration=moles/Vol. in litre molar concentration=(20/40)/5=0.1 Answer: (a)
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