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NEET-XII-Chemistry

02: Solutions

with Solutions - page 4
 
  • Qstn #26
    If
    the density of some lake water is 1.25 g mL-1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
    Ans : Number of moles present
    in 92 g of Na+ ions =
    = 4 mol
    Therefore, molality of
    Na+ ions in the lake
    = 4 m
  • Qstn #27
    If
    the solubility product of CuS is 6 × 10-16,
    calculate the maximum molarity of CuS in aqueous solution.
    Ans : Solubility product of
    CuS, Ksp = 6 × 10-16
    Let s be the
    solubility of CuS in mol L-1.

    Now,
    = s × s
    = s2
    Then, we have, Ksp =

    = 2.45 × 10-8 mol L-1
    Hence, the maximum
    molarity of CuS in an aqueous solution is 2.45 × 10-8 mol L-1.
  • Qstn #28
    Calculate
    the mass percentage of aspirin (C9H8O4)
    in acetonitrile (CH3CN)
    when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
    Ans : 6.5
    g of C9H8O4 is dissolved in 450 g of CH3CN.
    Then,
    total mass of the solution = (6.5 + 450) g
    = 456.5 g
    Therefore, mass
    percentage ofC9H8O4
    =
    1.424%
  • Qstn #29
    Nalorphene
    (C19H21NO3),
    similar to morphine, is used to combat withdrawal
    symptoms
    in narcotic users. Dose of nalorphene generally given is 1.5 mg.
    Calculate
    the mass of 1.5 × 10-3m
    aqueous solution required for the above dose.
    Ans : The
    molar mass of nalorphene is given as:

    In
    1.5 × 10-3m aqueous solution of nalorphene,
    1 kg
    (1000 g) of water contains 1.5 × 10-3 mol

    Therefore,
    total mass of the solution

    This
    implies that the mass of the solution
    containing 0.4665 g of nalorphene is 1000.4665 g.
    Therefore,
    mass of the solution containing 1.5 mg of nalorphene is:

    Hence, the
    mass of aqueous solution required is 3.22 g.
    Note: There is a slight variation in this answer and the one given in the
    NCERT textbook.
  • Qstn #30
    Calculate
    the amount of benzoic acid (C6H5COOH)
    required for preparing 250 mL of 0.15 M solution in methanol.
    Ans : 0.15 M solution of
    benzoic acid in methanol means,
    1000 mL of solution
    contains 0.15 mol of benzoic acid
    Therefore, 250 mL of
    solution contains = mol of benzoic acid
    = 0.0375 mol of
    benzoic acid
    Molar
    mass of benzoic acid (C6H5COOH)
    = 7 × 12 + 6 × 1 + 2 × 16
    =
    122 g mol-1
    Hence, required benzoic
    acid = 0.0375 mol × 122 g mol-1
    = 4.575 g
  • Qstn #31
    The
    depression in freezing point of water observed for the same amount of
    acetic acid, trichloroacetic acid and trifluoroacetic acid increases
    in the order given above. Explain briefly.
    Ans :
    Among H, Cl, and F, H
    is least electronegative while F is most electronegative. Then, F can
    withdraw electrons towards itself more than Cl and H. Thus,
    trifluoroacetic acid can easily lose H+ ions i.e.,
    trifluoroacetic acid ionizes to the largest extent. Now, the more
    ions produced, the greater is the depression of the freezing point.
    Hence, the depression in the freezing point increases in the order:
    Acetic acid <
    trichloroacetic acid < trifluoroacetic acid
  • Qstn #32
    Calculate
    the depression in the freezing point of water when 10 g of
    CH3CH2CHClCOOH
    is added to 250 g of water. Ka = 1.4 × 10-3, Kf = 1.86
    K kg
    mol-1.
    Ans : Molar
    mass of

    ∴No.
    of moles present in 10 g of

    It
    is given that 10 g of is
    added to 250 g of water.
    ∴Molality
    of the solution,

    Let α be the degree of dissociation of
    undergoes
    dissociation according to the following equation:



    Since α is very small with respect to 1, 1 - α
    ≈
    1
    Now,


    Again,

    Total
    moles of equilibrium = 1 - α
    + α
    + α
    = 1
    + α


    Hence, the
    depression in the freezing point of water is given as:

  • Qstn #33
    19.5
    g of CH2FCOOH
    is dissolved in 500 g of water. The depression in the freezing point
    of water observed is 1.0°C. Calculate the van’t Hoff factor
    and dissociation constant of fluoroacetic acid.
    Ans : It is
    given that:

    We know
    that:

    Therefore,
    observed molar mass of
    The
    calculated molar mass of is:

    Therefore,
    van’t Hoff factor,

    Let
    α
    be the degree of dissociation of


    Now,
    the value of Ka is given as:


    Taking
    the volume of the solution as 500 mL, we have the concentration:

    Therefore,

  • Qstn #34
    Vapour
    pressure of water at 293 Kis 17.535 mm Hg. Calculate the
    vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in
    450 g of water.
    Ans : Vapour pressure of
    water, =
    17.535 mm of Hg
    Mass of glucose, w2 = 25 g
    Mass of water, w1 = 450 g
    We know that,
    Molar mass of glucose
    (C6H12O6), M2 = 6
    × 12 + 12 × 1 + 6 × 16
    = 180 g mol-1
    Molar mass of water, M1 = 18 g mol-1
    Then, number of moles
    of glucose,
    = 0.139 mol
    And, number of moles of
    water,
    = 25 mol
    We know that,

    ⇒ 17.535 - p1 = 0.097
    ⇒ p1 = 17.44 mm of Hg
    Hence, the vapour
    pressure of water is 17.44 mm of Hg.
  • Qstn #35
    Henry’s
    law constant for the molality of methane in benzene at 298 Kis 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg.
    Ans : Here,
    p = 760 mm Hg
    kH = 4.27 ×
    105 mm Hg
    According to Henry’s
    law,
    p = kHx

    = 177.99 × 10-5
    = 178 × 10-5 (approximately)
    Hence, the mole
    fraction of methane in benzene is 178 × 10-5.
  • Qstn #36
    100
    g of liquid A (molar mass 140 g mol-1)
    was dissolved in 1000 g of liquid B (molar mass 180 g mol-1).
    The vapour pressure of pure liquid B was found to be 500 torr.
    Calculate the vapour pressure of pure liquid A and its vapour
    pressure in the solution if the total vapour pressure of the solution
    is 475 Torr.
    Ans : Number of moles of
    liquid A,
    = 0.714 mol
    Number of moles of
    liquid B,
    = 5.556 mol
    Then, mole fraction of
    A,

    = 0.114
    And, mole fraction of
    B, xB = 1 - 0.114
    = 0.886
    Vapour pressure of pure
    liquid B, =
    500 torr
    Therefore, vapour
    pressure of liquid B in the solution,

    = 500 × 0.886
    = 443 torr
    Total vapour pressure
    of the solution, ptotal = 475 torr
    Vapour
    pressure of liquid A in the solution,
    pA = ptotal - pB
    = 475 - 443
    = 32 torr
    Now,


    = 280.7 torr
    Hence, the vapour
    pressure of pure liquid A is 280.7 torr.
    Page No 62:
  • Qstn #37
    Vapour
    pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and
    632.8 mm Hg respectively. Assuming that they form ideal solution over
    the entire range of composition, plot ptotal’ pchloroform’
    and pacetone as a function of xacetone.
    The experimental data observed for different compositions of mixture
    is.
    100 ×xacetone 011.823.436.050.858.264.572.1
    pacetone /mm Hg054.9110.1202.4322.7405.9454.1521.1
    pchloroform/mm
    Hg    		632.8548.1469.4359.7257.7193.6161.2120.7
    Ans : From the
    question, we have the following data
    100 ×xacetone 011.823.436.050.858.264.572.1
    pacetone /mm Hg054.9110.1202.4322.7405.9454.1521.1
    pchloroform/mm
    Hg    		632.8548.1469.4359.7257.7193.6161.2120.7
    ptota(mm
    Hg)    		632.8603.0579.5562.1580.4599.5615.3641.8

    It
    can be observed from the graph that the plot for the ptotal of the solution curves downwards. Therefore, the solution shows
    negative deviation from the ideal behaviour.
  • Qstn #38
    Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
    Ans : Molar mass of benzene

    Molar mass of toluene

    Now, no. of moles present in 80 g of benzene
    And, no. of moles present in 100 g of toluene
    ∴Mole fraction of benzene, xb
    And, mole fraction of toluene,

    It is given that vapour pressure of pure benzene,

    And, vapour pressure of pure toluene,

    Therefore, partial vapour pressure of benzene,


    And, partial vapour pressure of toluene,


    Hence, mole fraction of benzene in vapour phase is given by:


  • Qstn #39
    The air is
    a mixture of a number of gases. The major components are oxygen
    and
    nitrogen with approximate proportion of 20% is to 79% by volume at
    298
    K.
    The water is in equilibrium with air at a pressure of 10 atm. At 298
    Kif the
    Henry’s
    law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
    Ans : Percentage
    of oxygen (O2)
    in air = 20 %
    Percentage
    of nitrogen (N2)
    in air = 79%
    Also, it
    is given that water is in equilibrium with air at a total pressure of
    10 atm, that is, (10 × 760) mm Hg = 7600 mm Hg
    Therefore,
    Partial
    pressure of oxygen,
    = 1520
    mm Hg
    Partial
    pressure of nitrogen,
    =
    6004 mmHg
    Now,
    according to Henry’s law:
    p = KH.x
    For
    oxygen:

    For
    nitrogen:

    Hence,
    the mole fractions of oxygen and nitrogen in water are 4.61 ×10-5and
    9.22 × 10-5 respectively.
  • Qstn #40
    Determine
    the amount of CaCl2 (i =
    2.47) dissolved in 2.5 litre of water such that its osmotic pressure
    is 0.75 atm at 27°C.
    Ans : We
    know that,


    Here,
    R
    = 0.0821 L atm K-1mol-1
    M
    = 1 × 40 + 2 × 35.5
    = 111g mol-1
    Therefore, w
    = 3.42 g
    Hence,
    the required amount of CaCl2 is 3.42 g.
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