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Solutions Test 14

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Solutions Test 14
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  • Question 1
    1 / -0

    Select correct statement -

    Solution

    (A) NaCl → Na+ + Cl–
    ΔTb = Kb × ms
    ΔTb = 2Kb × m
    elavation of b.p. will be double in case of NaCl not b.p.
    (B) Will be correct because b.p. elavation will be double here in comprasion to glucose.
    (C) Elavation of b.p. is colligative property not b.p

  • Question 2
    1 / -0

    Total vapour pressure of mixture of 1 mol volatile component A(pºA = 100 mmHg) and 3 mol of volatile component B (pºB = 60 mmHg) is 75 mm. For such case -

    Solution

    As, we know that

    = 25 + 45 mm Hg
    = 70 mm Hg
    To mm Hg < 75 mm Hg thus there is positive deviation from Raoults law,
    so (A) is correct
    (B) If v.p. is increasing boiling point will be lowered in that case
    (C) is correct that force of attraction between A & B is smaller than that between A and A or between B and B.
    So All statement are corrrect

  • Question 3
    1 / -0

    At a given temperature, total vapour pressure in Torr of a mixture of volatile components A and B is given by

    P = 120 – 75 XB

    hence, vapour pressure of pure A and B respectively (in Torr) are -

    Solution

    P = 120 – 75 × B
    If XB = 1
     = 120 – 70 = 45 mm Hg
    If XB = 120 – 75 = 45 mm Hg
    if  XB = 0    Ps =
    = 120 mm Hg

  • Question 4
    1 / -0

    Van't Hoff factors of aqueous solutions of X, Y, Z are 1.8, 0.8 and 2.5. Hence, their (assume equal concentrations in all three cases) then correct order is :-

    Solution

     

    As van't Holf factor increases RLVP increases i.e.,V,P. decreases y > x > z

  • Question 5
    1 / -0

    Decimolar solution of potassium ferricyanide, K3[Fe(CN)6] has osmotic pressure of 3.94 atm at 27ºC. Hence percent ionisation of the solute is -

    Solution

    π = iCRT
    K3[Fe(CN)6] = 3k+ + [Fe(CN)6]3–

    i = (1 + 3α)

    α = 0.2
    so  20%

  • Question 6
    1 / -0

    An aqueous solution of urea containing 18 g urea in 1500 cm3 of solution has a density of 1.052 g/cm3. If the molecular weight of urea is 60, then the molality of solution is-

    Solution

    The correct answer is Option B.

    Density is the ratio of mass to volume.
    Density is 1.052 g/ml.
    1500 ml of solution corresponds to       
    = 1.052 g/ml×1500ml
    = 1578 g     
    Molar mass of urea is 60 g/mol. Number of moles of urea is the ratio of mass to molar mass. It is 18g / 60g/mol = 0.3mol
     
    Molality of solution is the ratio of the number of moles of urea to the mass of solvent in kg.
    Mass of water =1578 g − 18 g
                              =1560g
                              =1.560kg (as 1 kg = 1000g)
    Molality = 0.3 / 1.560
                  = 0.192m
     

  • Question 7
    1 / -0

    2.56 g of sulphur in 100 g of CS2 has depression in f.p. of 0.010º, Kf = 0.1º (molal)-1. Hence, atomicity of sulphur is -

    Solution


    M = 256
    M = 8 × 32
    Thus atomicity = 8

  • Question 8
    1 / -0

    Consider following solutions -

    I : 1 M a glucose II. : 1 M a sodium chloride

    III. : 1 M benzoic acid in benzene IV. : 1 M ammonium phosphate

    Select incorrect statement -

    Solution

    (A) All solutions are not isotonic because all solution are not at equal concentration so (A) is incorrect
    (B) As benzoic acid dimerises so III is hypotonic of I, II, IV
    (C) Is also correct as I, II & IV are hypertonic of III
    (D) As ammoniun sulphate has maximum number of particale so it will be hypertonic of I, II, III

  • Question 9
    1 / -0

    The relationship between osmotic pressure at 273 K when 10 g glucose (P1) 10 g urea (P2) and 10 g sucrose (P3) are dissolved in 250 ml of water is -

    Solution

    As number of moles is maximum in case urea > glucose > sucrose
    π = CRT
    It depends on number of moles so osmotic pressure
    P2 > P1 > P3

  • Question 10
    1 / -0

    The osmotic pressure of a solution of benzoic acid dissolved in benzene is less than expected because-

    Solution

    As benzoic acid dimerises so number of moles decreases so osmotic pressure of benzoic acid is less than benzene.

  • Question 11
    1 / -0

    Assuming each salt to be 90% dissociated, which of the following will have highest osmotic pressure

    Solution

    Higher the number of particles higher will be osmoatic pressure so (A) will be answer in case of (D) precipitaiton will take place so number of particles will decrease.

  • Question 12
    1 / -0

    Which one of the following pairs of solution can we expect to be isotonic at the same temperature-

    Solution

    Isotonic means equal osmotic pressure so, π1 = π2
    i1C1RT = i2C2RT
    so i1C1 = i2C2
    In case of (D) both the volume of (i) & (C) are equal.

  • Question 13
    1 / -0

    For a solution containing non-volatile solute, the relative lowering of vapour pressure is 0.2. If the solution contains 5 moles in all, which of the following are true ?

    I. Mole fraction of solute in the solution is 0.2
    II. No. of moles of solute in the solution is 0.2
    III. No. of moles of solvent in the solution is 4
    IV. Mole fraction of solvent is 0.2 -

    Solution


    As we know relative lowering of v.p. is equal to the mole fraction of the solute so (I) is correct it doesn’t depend upon the number of moles so (II) is wrong mole fraction of solvent will be 0.8 so number of moles of solvent will be 4 so (III) will be correct, (IV) will be also wrong so I & III will be correct

  • Question 14
    1 / -0

    A complex containing K+, Pt (IV) and Cl- is 100% ionised giving i = 3. Thus, complex is

    Solution

    In option (B) oxidaton state of platinum is (iv)
    x – 6 = –2
    x = + 4

  • Question 15
    1 / -0

    If pKa = -log Ka = 4, and Ka = Cα2 then van't Hoff factor for weak monobasic acid when C = 0.01 M is-

    Solution

    i = 1 + α
    10–4 = Cα2

    i = 1 + 2 = 1 + 0.1 = 1.1

  • Question 16
    1 / -0

    pH of 1M HA (weak acid) is 2. Hence van't Hoff factor is -

    Solution

    HA = H+ + A–
    (1 – α) α       α
    PH = 2 = [H+]  : 10–2  = C2
    a = 0.1
    i = 1 + α = 1 + 0.1 = 1.01

  • Question 17
    1 / -0

    In which case van't Hoff factor is maximum

    Solution

    (a) As we know
     i = 1 + α

    i = 1 + α =  50 % dissociation
    i = 1.5

    1 + Zα ,  α = .4
    i = 1.8

    1 + 3α
    i = 1.9

    1 + 4α
    i = 1.8

  • Question 18
    1 / -0

    If 18 gram of glucose (C6H12O6) is present in 1000 gram of an aqueous solution of glucose it is said to be-

    Solution

    18 gm of glucose means 0.1mole of glucose as it present in 1000 gm of solvent so it is 0.1 mole

  • Question 19
    1 / -0

    What is the molarity of H2SO4 solution that has a density of 1.84 gm/cc at 350C and contains 98% by weight-

    Solution

  • Question 20
    1 / -0

    In order to prepare 100 cm3 of 0.250 M barium chloride solution the amount of BaCl2.2H2O required will be-

    Solution

    Molality of BaCl2 = 0.1 × 0.25 = 0.025
    by calculation we get the values of (D)

  • Question 21
    1 / -0

    25 mL of 3 M HCl were added to 75 mL of 0.05 M HCl. The molarity of HCl in the resulting solution is approximately-

    Solution

    Mole of HCl = 25 × 3 +75 × 0.05 = 75 + 3.75 = 78.75

  • Question 22
    1 / -0

    0.2 mole of HCl and 0.1 mole of CaCl2 were dissolved in water to have 500 ml of solution, the molarity of Cl¯ ions is-

    Solution

  • Question 23
    1 / -0

    When 5.0 gram of BaCl2 is dissolved in water to have 106 gram of solution. The concentration of solution is-

    Solution


    so concentration of solution = 5 ppm

  • Question 24
    1 / -0

    For an ideal binary liquid solution with , which relation between XA(molefraction of A in liquid phase) and YA (mole fraction of A in vapour phase) is correct?

    Solution



  • Question 25
    1 / -0

    Mole fraction of A vapours above the solution in mixture of A and B(XA = 0.4) will be

    [Given :  = 100 mm Hg and  = 200 mm Hg]

    Solution


    = 40 + 120 = 160

    π 40 = yA × 160
    ⇒ yA = ¼

  • Question 26
    1 / -0

    The exact mathematical expression of Raoult's law is

    Solution

    The exact mathematical expression of Raoult's law is 
    Here, Po represents the vapour pressure of the pure solvent, P represents the vapour pressure of the solution, n represents the number of moles of solute and N represents the number of moles of the solvent.

  • Question 27
    1 / -0

    A mixture contains 1 mole of volatile liquid A ( = 100 mm Hg) and 3 moles of volatile liquid B(= 80 mmHg). If solution behaves ideally, the total vapour pressure of the distillate is

    Solution




    = 85.88 mm Hg

  • Question 28
    1 / -0

    Which of the following aqueous solution will show maximum vapour pressure at 300 K?

    Solution

    less no of particle of solute means maximum vapour pressure.

  • Question 29
    1 / -0

    The Van't Hoff factor for a dilute aqueous solution of glucose is

    Solution

    Glucose does not dissociate
    ⇒ i = 1

  • Question 30
    1 / -0

    The correct relationship between the boiling points of very dilute solution of AlCl3(T1K) and CaCl2(T2K) having the same molar concentration is

    Solution

    AlCl3 = no. of particle = 4
    CaCl2 = no. of particle = 3
    (vapour pressure of AlCl3) < (vapour pressure of CaCl2 solution)
    TB.P. (AlCl3) > TB.P. (CaCl2)
    T1 > T2

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