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NEET Chemistry MCQ
Solutions And Liquid States Mcq Neet Chemistry
Quiz 3
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Q.1
Density of a 2.05M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is [ AIEEE2006]
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a) 1.14 mol/Kg
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b) 3.28 mol/Kg
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c) 2.28 mol /kg
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d) 0.44 mol/kg
Explanation
M = molarity, d = density , M2 = molecular mass , m = molality Answer: (c)
Q.2
During the evaporation of liquid
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a) The temperature remains unaffected
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b) May rise or fall depending on the nature
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c) The temperature of the liquid will fall
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d) The temperature of the liquid will rise
Explanation
Answer:(c)
Q.3
18g of glucose (C6H12O6) is added to 178.2 g of water. the vapour pressure of water for this aqueous solution at 100°C is [ AIEEE 2006]
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a) 759.00 torr
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b) 7.60 torr
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c) 76.00 torr
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d) 752.40 torr
Explanation
Vapour pressure of water at 100°C is 760 torr From the formula for relative depression of vapour pressure Answer: (d)
Q.4
A mixture of ethyl alcohol and propyl alcohol has vapour pressure of 290 mm at 300K. The vapour pressure of propyl alcohol is 200 nm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure ( in mm) at the same temperature will be [ AIEEE 2007]
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a) 360
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b) 350
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c) 300
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d) 700
Explanation
According to Raoult's law p = pA + pB Answer: (b)
Q.5
Which has minimum osmotic pressure?
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a) 200 ml of 2M NaCl solution
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b) 200 ml of 1M glucose solution
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c) 200 ml of 2M urea solution
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d) All have same
Explanation
Osmotic pressure is colligative property which depends upon the number of particles Concentration of particles is least in 1 M glucose solution. So 1M glucose solution has minimum osmotic pressure Answer: (b)
Q.6
If 0.5m of Ca(NO3)2 and 0.75m of KOH is taken , then the depression in freezing point is
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a) Greater in Ca(NO3)2 because number of ions are greater
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b) Greater in KOH because concentration is high
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c) Equal in both and freezing point is less than 0°C because ionic concentration is same
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d) Equal to 0°C in both because ionic concentration is negligible
Explanation
Ionic concentration of Ca(NO3)2 = 0.5 ×3 = 1.5 Ionic concentration of KOH= = 0.75 ×2 = 1.5 Ionic concentration is same so depression in freezing point will also be same Answer:(c)
Q.7
The osmotic pressure of equimolar solution of urea, BaCl2 and AlCl3 will be in the order
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a)BaCl2 > Urea > AlCl3
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b) Urea > BaCl2 > AlCl3
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c) BaCl2 > AlCl3 > Urea
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d) AlCl3 > BaCl2 > Urea
Explanation
According to van't Hoff equation π = iCRT → π ∝ i , when concentration (C) is constant ( for equimolar solutions) ∴ i for urea = 1 i for BaCl2 = 3 i for AlCl3 = 4 More the value of "i" more is the pressure thus "d" is correct option Answer: (d)
Q.8
Which will form maximum boiling point azeotrope
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a) HNO3 + H2O solution
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b) C2H5OH + H2O solution
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c) C6H6 + C6H5CH3 solution
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d) None of these
Explanation
HNO3 + H2O solution will form maximum boiling point azeotrope because they have a composition of having maximum boiling point Answer: (a)
Q.9
When 20 g of naphthoic acid ( C11H5O2) is dissolved in 50 g of benzene ( Kf = 1.72 K Kg mole -1), a freezing point depression of 2K is observed. the van't Hoff factor (i) is
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a) 0.5
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b) 1
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c) 2
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d) 3
Explanation
For electrolytic solution ΔTf = iKfm Answer: (a)
Q.10
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea ( molar mass = 60 g/mol) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0g/cm3, molar mass of the substance will be [ AIEEE 2007]
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a) 210.0 g/mol
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b) 90.0 g/mol
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c) 115 g/mol
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d) 105.0 g/mol
Explanation
Isotonic solutions have same osmotic pressure π1 = C1RT π2 = C2RT For isotonic solutions π1 = π2 ∴ C1 = C2 Answer:(a)
Q.11
The freezing point of equimolal aqueous solution will be highest for
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a) Glucose
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b) La(NO3)2
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c) Ca(NO3)2
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d) C6H5NH3Cl
Explanation
ΔTf = iKfm since Kf and m are constant ΔKf ∝ i i is lowest for glucose ( i =1) and the lower the value of Δf, higher is the freezing point Answer: (a)
Q.12
Which has highest freezing point?
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a) 1 m K4[Fe(CN)6] solution
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b) 1 m C6H12O6 solution
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c) 1 m KCl solution
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d) 1 m rock salt solution
Explanation
A substance which in the aqueous solution furnishes the minimum number of particles will lower the freezing point to the minimum. Therefore, this solution will have the maximum freezing point C6H12O6, being a non-electrolyte, furnishes minimum number of particles among the given four in the solution Answer: (b)
Q.13
At 80°, the vapour pressure of pure liquid A is 520mm of Hg and that of pure liquid B is 1000 mm of Hg. If a mixture solution of A and B boils at 80°C and 1atm pressure, the amount of A in the mixture is ( 1 atm = 760 mm of Hg)
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a) 50 mol percent
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b) 52 mol percent
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c) 34 mol percent
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d) 48 mol percent
Explanation
From formula or 50 mol percent Answer: (a)
Q.14
X is dissolved in water. Maximum boiling point is observed when X is ... ( 0.1 M each)
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a) CaSO4
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b) BaCl2
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c) NaCl
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d) Urea
Explanation
Elevation in boiling point is a colligative property which depends upon number of molecules/ions. Higher the number of ions, greater will be elevation in boiling point one mole of BaCl2 will produce 3 moles of ions, which is highest in given option Answer:(b)
Q.15
During depression in freezing point of a solution the following are in equilibrium [ IIT 2003]
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a) Liquid solvent, solid solvent
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b) Liquid solvent, solid solute
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c) Liquid solute, solid solute
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d) Liquid solute, solid solvent
Explanation
Both liquid and solid phases of solvent are at equilibrium and both have same vapour pressure Answer: (a)
Q.16
Which of the following solution will have the highest boiling point? .. [ CPMT 1991]
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a) 1% solution of glucose in water
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b) 1% solution of sucrose in water
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c) 1% solution of sodium chloride in water
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d) 1% solution of calcium chloride in water
Explanation
Molar concentration Glucose=10/180MSucrose=10/342 MNaCl=(10/58.5) ×2 [ as two ions are produce from one molecule]CaCl2=(10/111) ×3 M Highest concentration of particle is in NaClAnswer: (c)
Q.17
Two liquids X and Y form an ideal solution. At 300K vapour pressure of the solution containing 1 mol of X and 3mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure ( in mm Hg) of X and Y in their pure states will be, respectively [ AIEEE 2009]
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a) 200 and 300
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b) 300 and 400
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c) 400 and 600
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d) 500 and 600
Explanation
On solving equation (1) and (2) we get pxo = 600 mm Hg and pyo = 400 mm Hg Answer: (c)
Q.18
The molar freezing point constant for water is 1.86°C/mole. If 342gm of cane sugar ( C12H22O11) is dissolved in 1000g of water, the solution will freeze at
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a) -3.92°C
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b) 1.86°C
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c) -1.86°C
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d) 2.42°C
Explanation
Kf = 1.86°C/mole molality of solution m ΔTf = Kf × m = 1.86 × 1 = 1.86 °C Now Tf = To - ΔTf Tf = 0 -1.86 = -1.86°C Answer: (c)
Q.19
Vapour pressure of solution of 5g of no-electrolte in 100 g of water at a particular temperature is 2985 N/mThe vapour pressure of pure water is 3000 N/m2, the molecular weight of the solution is [ IIT 1993]
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a) 90
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b) 200
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c)180
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d)380
Explanation
M2=molar mass of solute w2=weight of solute M1=Molar mass of solvent w1 weight of solvent po=vapour pressure of pure solvent p=vapour pressure of solution Answer:(c)
Q.20
A 0.6% urea solution is isotonic with
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a) 0.6% NaCl solution
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b) 0.6% glucose solution
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c) 0.1 M KCl solution
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d) 0.1 M glucose solution
Explanation
0.6% urea and 0.6% glucose are isotonic as both of them do not undergo ionisation Answer: (b)
Q.21
The henry's law constant for the solubility of N2 gas in water at 298K is 1.0×105 atm. The mole fraction of N2 in air is 0.the number of moles of N2 from air dissolved in 10 moles of water at 298K and 5atm pressure is [ IIT 2009]
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a) 4.0 × 10-4
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b) 4.0 × 10-5
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c) 5.0 × 10-4
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d) 4.0 × 10-6
Explanation
According to Henry's law Given total pressure=5 atm mole fraction of N2=0.5 ∴ partial pressure of N2=0.8 ×5=4 H × x Here p=partial pressure of Nitrogen) ∴ 4=1.0×105 × X X=4×10-5 number of moles of H2O N=10 given Number of moles of N2 (n)=? Answer: (a)
Q.22
The vapour pressure of a solvant decreased by 10 mm of mercury when non-volatile solute was added to the solvant. the mole fraction of the solute in the solution is 0.What should be the mole fraction of the solvant, if the decrase in the vapour pressure is to be 20mm of mercury
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a) 0.8
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b) 0.6
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c)0.4
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d)0.2
Explanation
Mole fraction of solute Comparing under the two conditions Mole freaction of solvanr=1-0.4=0.6 Answer: (b)
Q.23
A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behavior of the solution? [ AIEEE 2009]
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a)The solution formed is an ideal solution
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b) The solution is non-ideal, showing +Ve deviation from Raoult's law
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c)The solution is non-ideal, showing -Ve deviation from Raoult's law
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d)n-heptane shows +Ve deviation while ethanol shows -Ve deviation from Raoult's law [ AIEEE 2009]
Explanation
The solution containing n-heptane and ethanol shows non-deal behavior with positive deviation from Raoult's law. this is because the ethanol molecules are held together by strong H-bonds, however the forces between n-heptane and ethanol are not very strong, as a result they easily vapourised showing higher vapour pressure than expected Answer:(b)
Q.24
If molecular weight of a compound is increased then sensitivity is decreased in which of the following methods
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a) Dialysis
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b) Osmosis
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c) Viscosity
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d) Elevation in boiling point
Explanation
All colligative properties have inverse relationship with molecular weight of the solute Answer: (d)
Q.25
Equimolal solutions in the same solvent have [ AIEEE 2005]
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a)Same boiling point but different freezing point
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b) Same freezing point but different boiling point
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c)Same boiling point and same freezing points
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d)Different boiling point and different freezing points
Explanation
given solutions are equimolal thus have same boiling point and same freezing pointsAnswer: (c)
Q.26
The density (in g mL-1) of a 3.60M sulphuric acid solution that is 29% H2SO4 ( molar mass=98 g/mol) by mass will be [ AIEEE 2007]
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a) 1.45
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b) 1.64
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c)1.88
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d)1.22
Explanation
3.6 M solution means 3.6 moles of H2SO4 is present in 1000 ml of solution ∴ Mass of 3.6 moles of H2SO4=3.6 × 98 g=352.8 g ∴ Mass of H2SO4 in 1000 ml of solution=352.8 g Given, 29g of H2SO4 is present in 100g of solution ∴ 352.8 g of H2SO4 is present in(100/29) × 352.8=1216 g of solution Now, density=Mass/ Volume=1216 / 1000=1.216 g/mol=1.22 g/mlAnswer: (d)
Q.27
Solution A contains 7g/L of MgCl2 and solution B contains 7g/L of NaCl. At room temperature, the osmotic pressure of
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a) Solution A is greater than B
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b) Both have same osmotic pressure
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c)Solution B is greater than A
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d)Can't determine
Explanation
The osmotic pressure of solution A is greater than B due to greater number of particles Answer:(a)
Q.28
60g of urea ( Mol wt=60) was dissolved in 9.9 moles water. If the vapour pressure of pure water is p°, the relative lowering in vapour pressure of the solution is
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a) 0.87
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b) 0.09
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c) 0.99
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d) 0.10
Explanation
Number of moles of solute ( urea)=60/60=1 Relative lowering in vapour pressure Where p° is vapour pressure of solvent p=vapour pressure of solution Xsolute=mole fraction of soluteAnswer: (b)
Q.29
Which of the following compounds corresponds to van't Hoff factor (i) to be equal to 2 for dilute solution? .. [ NCERT 1978]
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a) K2SO4
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b) NaHSO4
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c) sugar
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d) MgSO4
Explanation
MgSO4 dissociate to give two ions Answer: (d)
Q.30
When a substance is dissolved in a solvent the results in .... [ NCERT 1981]
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a) an increase in the b.p. of the solution
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b) decrease in the b.p. of the solvent
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c) the solution having higher freezing point than the solvent
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d) the solution having a lower osmotic pressure than the solvent
Explanation
Answer: (a)
0 h : 0 m : 1 s
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