JEE Questions for Chemistry Solutions Quiz 5 - MCQExams.com

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?
  • The solution formed is an ideal solution
  • The solution is non-ideal, showing positive deviation from Raoult's law
  • The solution is non-ideal, showing negative deviation from Raoult's law
  • n-heptane shows positive deviation while ethanol show negative deviation from Raoult's law
Vapour pressure of pure 'A' is 70 mm of Hg at 25°C. It forms an ideal solution with 'B ' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25°C, the vapour pressure of pure `B ' at 25°C is
  • 28 mm
  • 56 mm
  • 70 mm
  • 140 mm
The vapour pressure of two liquids X and Y are 80 and 60 Torr respectively. The total vapour pressure of the ideal solution obtained by mixing 3 moles of X and 2 moles of Y would be
  • 68 torr
  • 140 torr
  • 48 torr
  • 72 torr
  • 54 torr
One component of a solution follows Raoult\'s law over the entire range 0 ≤ x1 ≤ 1. The second component must follow Raoult\'s law in the range when x2 is
  • close to zero
  • close to 1
  • 0 ≤ x2 ≤ 0.5
  • 0 ≤ x2 ≤ 1
At 80°C, the vapour pressure of pure liquid 'A ' is 520 mm Hg and that of pure liquid 'B ' is 1000 mm Hg. If a mixture solution of `A ' and `B ' boils at 80°C and 1 atm pressure, the amount of 'A ' in the mixture is (1 atm = 760 mm Hg)
  • 52 mole per cent
  • 34 mole per cent
  • 48 mole per cent
  • 50 mole per cent
Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1: 1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are
  • 250 mm, 550 mm
  • 350 mm, 450 mm
  • 350 mm, 700 mm
  • 500 mm, 500 mm
  • 550 mm, 250 mm
Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is

  • Chemistry-Solutions-7013.png
  • 2)
    Chemistry-Solutions-7014.png

  • Chemistry-Solutions-7015.png

  • Chemistry-Solutions-7016.png
Formation of a solution from two components can be considered as
I pure solvent→ separated solvent molecules, ∆H1
II pure solute→ separated solute molecules, ∆H2
III separated solvent and solute molecules → solution, ∆H3
Solution so formed will be ideal if
  • ∆Hsoln = ∆H1 - ∆H2 - ∆H3
  • ∆Hsoln = ∆H3 - ∆H1 - ∆H2
  • ∆Hsoln = ∆H1 + ∆H2 + ∆H3
  • ∆Hsoln = ∆H1 + ∆H2 - ∆H3
On a humid day in summer, the mole fraction of gaseous H2O (water vapour) in the air at 25°C can be as high as 0.0287. Assuming a total pressure of 0.977 atm. What is the partial pressure of dry air ?
  • 94.9 atm
  • 0.949 atm
  • 949 atm
  • 0.648 atm
  • 1.248 atm
At 25°C, the total pressure of an ideal solution obtained by mixing 3 moles of 'A ' and 2 moles of 'B ', is 184 torr. What is the vapour pressure (in torr) of pure ' B ' at the same temperature ? (Vapour pressure of pure 'A' at 25°C is 200 torr)
  • 180
  • 160
  • 16
  • 100
A and B are ideal gases. The molecular weights of A and B are in the ratio of 1: 4. The pressure of a gas mixture containing equal weights of A and B is p atm. What is the partial pressure (in atm) of B in the mixture ?

  • Chemistry-Solutions-7017.png
  • 2)
    Chemistry-Solutions-7018.png

  • Chemistry-Solutions-7019.png

  • Chemistry-Solutions-7020.png
In a mixture of A and B, components show negative deviation when
  • A—B interaction is stronger than A —A and B —B interaction
  • A —B interaction is weaker than A— A and B — B interaction
  • ∆Vmix > 0, ∆Smix > 0
  • ∆Vmix = 0, ∆Smix > 0
If the elevation in boiling point of a solution of 10 g of solute (mol. wt =in 100 g of water is ∆Tb , the ebullioscopic constant of water is
  • 10
  • 100 Tb
  • ∆Tb
  • ∆Tb ⁄10
If the osmotic pressure of a 0.010 M aqueous solution of sucrose at 27° C is 0.25 atm, then the osmotic pressure of a 0.010 M aqueous solution of NaCl at 27° C is
  • 0.062 atm
  • 0.12 atm
  • 0.25 atm
  • 0.50 atm
A solution of 1.25 g of P in 50 g of water lower freezing point by 0.3° C. Molar mass of P is 94, Kf (water) =1.86 K kg mol-1 . The degree of association of P in water is
  • 80%
  • 60%
  • 65%
  • 75%
Glucose is mixed in 1 litre water, the ratio of ∆Tf / Kf changed to 10-3. The mass of glucose (C6H12O6 ) is
  • 180 gram
  • 18 gram
  • 1.8 gram
  • 0.18 gram
The molar mass of a solute X in g mol-1, if its 1% solution is isotonic with a 5% solution of cane sugar (molar mass = 342 g mol-1) is
  • 68.4
  • 34.2
  • 136.2
  • 171.2
Find the boiling point of a solution of 5.00 g of napthalene (C10 H8 ) in 100 g of benzene. Kb of benzene if 2.53°c/ m the normal boiling point of benzene = 80° C.
  • 81°C
  • 85°C
  • 0.9°C
  • 79°C
45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water, what is depression in freezing point?
  • 7.9 K
  • 2.5 K
  • 6.6 K
  • 2.2 K
An aqueous solution freezes at - 0.186° C then elevation in boiling point is (Kb = 0.512, Kf =1.
  • 0.0512° C
  • 100.0512° C
  • - 0.0512° C
  • None of these
Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2 H6 O2 ) must you add to get the freezing point of the solution lowered to - 2.8° C?
  • 72 g
  • 93 g
  • 39 g
  • 27 g
The mass of a non-volatile solute of molar mass 40 g mol-1 that should dissolved in 114 g of octane to lower its vapour pressure by 20%, is
  • 11.4 g
  • 9.8 g
  • 12.8 g
  • 10 g
58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point (b.p.) of the resulting solutions.
  • NaCl solution will show higher elevation of boiling
  • Glucose solution will show higher elevation of boiling point
  • Both the solutions will show equal elevation of boiling point.
  • The boiling point elevation will be shown by neither of the solutions.
The relative lowering of vapour pressure of a dilute solution is equal to mole fraction of solute present in the solution. Which of the following law is state this?
  • Henry law
  • Avogadro's law
  • Raoult's law
  • Law of definite proportion
The correct order of increasing boiling points of the following aqueous solutions
0.0001 M NaCl(I), 0.0001 M Urea(II), 0.001 M MgCl2(III), 0.01 M NaCl (IV) is
  • I < II < III < IV
  • IV < III < II < I
  • II < I < III < IV
  • III < II < IV < I
The freezing point (in °C) of solution containing 0.1 g of K3 [Fe(CN)6] (mol. wt.in 100 g of water (Kf =1.86 K kg mol-1) is
  • - 2.3 × 10-2
  • - 5.7 × 10-2
  • - 5.7 × 10-3
  • -1.2 × 10-2
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at - 6°C will be (Kf for water = 1.86 K kg mol-1 and molar mass of ethylene glycol = 62 g mol-1)
  • 804.32 g
  • 204.30 g
  • 400.00 g
  • 304.60 g
The empirical formula of a non-electrolyte is CH2O. A solution containing 3 g of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecular formula of the compound is
  • CH2O
  • C2H4O2
  • C4H8O4
  • C3H6O3
At 25°C a 5% aqueous solution of glucose (molecular weight = 180 g mol-1) is isotonic with 2% aqueous solution containing an unknown solute. What is the molecular weight of the unknown solute.
  • 60
  • 80
  • 98
  • 72
  • 63
A solution containing 1.8 g of a compound (empirical formula CH2O) in 40 g of water is observed to freeze at -0.465°C. The molecular formula of the compound is (Kf of water =1.86 kg K mol-1)
  • C2H4O2
  • C3H6O3
  • C4H8O4
  • C5H10O5
  • C6H12O6
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