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NEET-XII-Chemistry

02: Solutions

with Solutions - page 2

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  • Qstn #9
    Vapour
    pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea
    (NH2CONH2)
    is dissolved in 850 g of water. Calculate the vapour pressure of
    water for this solution and its relative lowering.
    Ans : It
    is given that vapour pressure of water, =
    23.8 mm of Hg
    Weight
    of water taken, w1 = 850 g
    Weight
    of urea taken, w2 = 50 g
    Molecular
    weight of water, M1 = 18 g mol-1
    Molecular
    weight of urea, M2 = 60 g mol-1
    Now,
    we have to calculate vapour pressure of water in the solution. We
    take vapour pressure as p1.
    Now,
    from Raoult’s law, we have:

    Hence,
    the vapour pressure of water in the given solution is 23.4 mm of Hg
    and its relative lowering is 0.0173.
  • Qstn #10
    Boiling
    point of water at 750 mm Hg is 99.63°C. How much sucrose is to be
    added to 500 g of water such that it boils at 100°C.
    Molal elevation constant for water is 0.52 K kg mol-1.
    Ans : Here,
    elevation of boiling point ΔTb = (100 + 273) - (99.63 + 273)
    =
    0.37 K
    Mass
    of water, wl = 500 g
    Molar
    mass of sucrose (C12H22O11), M2 = 11 × 12 + 22 × 1 + 11 × 16
    =
    342 g mol-1
    Molal
    elevation constant, Kb = 0.52 K kg mol-1
    We
    know that:


    =
    121.67 g (approximately)
    Hence,
    121.67 g of sucrose is to be added.
    Note: There is a slight variation in this answer and the one given in the
    NCERT textbook.
  • Qstn #11
    Calculate
    the mass of ascorbic acid (Vitamin C, C6H8O6)
    to be dissolved in
    75 g of acetic acid to lower its melting point
    by 1.5°C. Kf = 3.9 K kg mol-1.
    Ans : Mass
    of acetic acid, w1 = 75 g
    Molar
    mass of ascorbic acid (C6H8O6), M2 = 6 × 12 + 8 × 1 + 6 × 16
    =
    176 g mol-1
    Lowering
    of melting point, ΔTf = 1.5 K
    We
    know that:


    = 5.08 g
    (approx)
    Hence,
    5.08 g of ascorbic acid is needed to be dissolved.
    Note: There is a slight variation in this answer and the one given in the
    NCERT textbook.
  • Qstn #12
    Calculate
    the osmotic pressure in pascals exerted by a solution prepared by
    dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water
    at 37°C.
    Ans : It is
    given that:
    Volume
    of water, V = 450 mL = 0.45 L
    Temperature, T = (37 +
    273)K = 310 K
    Number
    of moles of the polymer,
    We know
    that:
    Osmotic
    pressure,

    = 30.98 Pa
    = 31
    Pa (approximately)
  • #
    Section : B
    Page No 59:
  • Qstn #1
    Define
    the term solution. How many types of solutions are formed? Write
    briefly about each type with an example.
    Ans : Homogeneous mixtures of
    two or more than two components are known as solutions.
    There are three types
    of solutions.
    (i) Gaseous
    solution:
    The
    solution in which the solvent is a gas is called a gaseous solution.
    In these solutions, the solute may be liquid, solid, or gas. For
    example, a mixture of oxygen and nitrogen gas is a gaseous solution.
    (ii) Liquid
    solution:
    The
    solution in which the solvent is a liquid is known as a liquid
    solution. The solute in these solutions may be gas, liquid, or solid.
    For
    example, a solution of ethanol in water is a liquid solution.
    (iii) Solid
    solution:
    The
    solution in which the solvent is a solid is known as a solid
    solution. The solute may be gas, liquid or solid. For example, a
    solution of copper in gold is a solid solution.
  • Qstn #2
    Give
    an example of solid solution in which the solute is a gas.
    Ans : In case a
    solid solution is formed between two substances (one having very
    large particles and the other having very small particles), an
    interstitial solid solution will be formed. For example, a solution
    of hydrogen in palladium is a solid solution in which the solute is a
    gas.
  • Qstn #3
    Define
    the following terms:
    (i) Mole fraction
    (ii) Molality
    (iii) Molarity
    (iv) Mass percentage.
    Ans : (i) Mole fraction:
    The
    mole fraction of a component in a mixture is defined as the ratio of
    the number of moles of the component to the total number of moles of
    all the components in the mixture.
    i.e.,
    Mole
    fraction of a component
    Mole
    fraction is denoted by ‘x’.
    If
    in a binary solution, the number of moles of the solute and the
    solvent are nA and nB respectively, then the mole fraction of the solute in the solution is
    given by,

    Similarly,
    the mole fraction of the solvent in the solution is given as:

    (ii) Molality
    Molality
    (m) is defined as the number of moles of the solute per kilogram of
    the solvent. It is expressed as:
    Molality
    (m)
    (iii) Molarity
    Molarity
    (M) is defined as the number of moles of the solute dissolved in one
    Litre of the solution.
    It
    is expressed as:
    Molarity
    (M)
    (iv) Mass
    percentage:
    The
    mass percentage of a component of a solution is defined as the mass
    of the solute in grams present in 100 g of the solution. It is
    expressed as:
    Mass
    % of a component
  • Qstn #4
    Concentrated
    nitric acid used in laboratory work is 68% nitric acid by mass in
    aqueous solution. What should be the molarity of such a sample of the
    acid if the density of the solution is 1.504 g mL-1?
    Ans : Concentrated nitric
    acid used in laboratory work is 68% nitric acid by mass in an aqueous
    solution. This means that 68 g of nitric acid is dissolved in 100 g
    of the solution.
    Molar mass of nitric
    acid (HNO3) = 1 ×
    1 + 1 × 14 + 3 ×
    16 = 63 g mol-1
    Then, number of moles
    of HNO3

    Given,
    Density of solution =
    1.504 g mL-1
    Volume
    of 100 g solution =

    Molarity of solution

    Page No 60:
  • Qstn #5
    A
    solution of glucose in water is labelled as 10% w/w, what would be
    the molality and mole fraction of each component in the solution? If
    the density of solution is 1.2 g mL-1,
    then what shall be the molarity of the solution?
    Ans : 10% w/w solution of
    glucose in water means that 10 g of glucose in present in 100 g of
    the solution i.e., 10 g of glucose is present in (100 - 10) g =
    90 g of water.
    Molar mass of glucose
    (C6H12O6) = 6 ×
    12 + 12 × 1 + 6 ×
    16 = 180 g mol-1
    Then, number of moles
    of glucose
    =
    0.056 mol
    Molality
    of solution =
    0.62 m
    Number of moles of
    water
    = 5 mol
    Mole
    fraction of glucose
    And, mole fraction of
    water
    = 1 - 0.011
    = 0.989
    If the density of the
    solution is 1.2 g mL-1, then the volume of the 100 g
    solution can be given as:

    Molarity
    of the solution
    = 0.67 M
  • Qstn #6
    How
    many mL of 0.1 M HCl are required to react completely with 1 g
    mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
    Ans : Let the amount of
    Na2CO3 in the mixture be x g.
    Then, the amount of
    NaHCO3 in the mixture is (1 - x) g.
    Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16
    = 106 g mol-1
    Number of moles Na2CO3
    Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16
    = 84 g mol-1
    Number
    of moles of NaHCO3
    According to the
    question,

    ⇒ 84x =
    106 - 106x
    ⇒ 190x =
    106
    ⇒ x =
    0.5579
    Therefore, number of
    moles of Na2CO3
    = 0.0053 mol
    And, number of moles of
    NaHCO3
    = 0.0053 mol
    HCl reacts with Na2CO3 and NaHCO3 according to the following equation.


    1 mol of Na2CO3 reacts with 2 mol of HCl.
    Therefore, 0.0053 mol
    of Na2CO3 reacts with 2 × 0.0053 mol =
    0.0106 mol.
    Similarly, 1 mol of
    NaHCO3 reacts with 1 mol of HCl.
    Therefore, 0.0053 mol
    of NaHCO3 reacts with 0.0053 mol of HCl.
    Total moles of HCl
    required = (0.0106 + 0.0053) mol
    = 0.0159 mol
    In 0.1 M of HCl,
    0.1 mol of HCl is
    preset in 1000 mL of the solution.
    Therefore, 0.0159 mol
    of HCl is present in
    = 159 mL of the
    solution
    Hence, 159 mL of 0.1 M
    of HCl is required to react completely with 1 g
    mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both.
  • Qstn #7
    A
    solution is obtained by mixing 300 g of 25% solution and 400 g of 40%
    solution by mass. Calculate the mass percentage of the resulting
    solution.
    Ans : Total amount of solute
    present in the mixture is given by,

    = 75 + 160
    = 235 g
    Total amount of
    solution = 300 + 400 = 700 g
    Therefore, mass
    percentage (w/w) of the solute in the resulting solution,
    = 33.57%
    And, mass percentage
    (w/w) of the solvent in the resulting solution,
    = (100 - 33.57)%
    = 66.43%
  • Qstn #8
    An
    antifreeze solution is prepared from 222.6 g of ethylene glycol
    (C2H6O2)
    and 200 g of water. Calculate the molality of the solution. If the
    density of the solution is 1.072 g mL-1,
    then what shall be the molarity of the solution?
    Ans : Molar mass of ethylene
    glycol=
    2 × 12 + 6 × 1 + 2 ×16
    = 62 gmol-1
    Number of moles of
    ethylene glycol
    = 3.59 mol
    Therefore, molality of
    the solution
    = 17.95 m
    Total mass of the
    solution = (222.6 + 200) g
    = 422.6 g
    Given,
    Density of the solution
    = 1.072 g mL-1
    Volume
    of the solution
    = 394.22 mL
    = 0.3942 × 10-3 L
    Molarity of the solution
    = 9.11 M
  • Qstn #9
    A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
    (i) express this in percent by mass
    (ii) determine the molality of chloroform in the water sample.
    digAnsr:   b
    Ans : (i) 15 ppm (by mass) means 15 parts per million (106) of the solution.
    Therefore, percent by mass
    = 1.5 × 10-3 %
    (ii) Molar mass of chloroform (CHCl3) = 1 × 12 + 1 × 1 + 3 × 35.5
    = 119.5 g mol-1
    Now, according to the question,
    15 g of chloroform is present in 106 g of the solution.
    i.e., 15 g of chloroform is present in (106 - 15) ≈ 106 g of water.
    Molality of the solution
    = 1.26 × 10-4 m
  • Qstn #10
    What
    role does the molecular interaction play in a solution of alcohol and
    water?
    Ans : In pure alcohol and
    water, the molecules are held tightly by a strong hydrogen bonding.
    The interaction between the molecules of alcohol and water is weaker
    than alcohol-alcohol and water-water interactions. As a
    result, when alcohol and water are mixed, the intermolecular
    interactions become weaker and the molecules can easily escape. This
    increases the vapour pressure of the solution, which in turn lowers
    the boiling point of the resulting solution.
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