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NEET-XII-Chemistry

02: Solutions

with Solutions - page 3

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  • Qstn #11
    Why
    do gases always tend to be less soluble in liquids as the temperature
    is raised?
    Ans : Solubility of gases in
    liquids decreases with an increase in temperature. This is because
    dissolution of gases in liquids is an exothermic process.

    Therefore, when the
    temperature is increased, heat is supplied and the equilibrium shifts
    backwards, thereby decreasing the solubility of gases.
  • Qstn #12
    State
    Henry’s law and mention some important applications?
    Ans : Henry’s law
    states that partial pressure of a gas in the vapour phase is
    proportional to the mole fraction of the gas in the solution.     If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be
    expressed as:
    p = KH x
    Where,
    KH is
    Henry’s law constant
    Some important
    applications of Henry’s law are mentioned below.
    (i) Bottles are sealed under high pressure to increase the
    solubility of CO2 in soft drinks and soda water.
    (ii) Henry’s law states that the solubility of gases
    increases with an increase in pressure. Therefore, when a scuba diver
    dives deep into the sea, the increased sea pressure causes the
    nitrogen present in air to dissolve in his blood in great amounts. As
    a result, when he comes back to the surface, the solubility of
    nitrogen again decreases and the dissolved gas is released, leading
    to the formation of nitrogen bubbles in the blood. This results in
    the blockage of capillaries and leads to a medical condition known as
    ‘bends’ or ‘decompression sickness’.
    Hence,
    the oxygen tanks used by scuba divers are filled with air and diluted
    with helium to avoid bends.
    (iii) The concentration of oxygen is low in the blood and
    tissues of people living at high altitudes such as climbers. This is
    because at high altitudes, partial pressure of oxygen is less than
    that at ground level. Low-blood oxygen causes climbers to become weak
    and disables them from thinking clearly. These are symptoms of
    anoxia.
  • Qstn #13
    The partial pressure of ethane over a solution containing 6.56 × 10-3 g of ethane is 1 bar. If the solution contains 5.00 × 10-2 g of ethane, then what shall be the partial pressure of the gas?
    Ans : Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1
    = 30 g mol-1
    Number of moles present in 6.56 × 10-3 g of ethane
    = 2.187 × 10-4 mol
    Let the number of moles of the solvent be x.
    According to Henry’s law,
    p = KHx

    Number of moles present in 5.00 × 10-2 g of ethane
    = 1.67 × 10-3 mol
    According to Henry’s law,
    p = KHx

    = 7.636 bar
    Hence, partial pressure of the gas shall be 7.636 bar.
  • Qstn #14
    What
    is meant by positive and negative deviations from Raoult’s law and
    how is the sign of ΔsolH related to positive and negative deviations from Raoult’s law?
    Ans : According
    to Raoult’s law, the partial vapour pressure of each volatile
    component in any solution is directly proportional to its mole
    fraction. The solutions which obey Raoult’s law over the entire
    range of concentration are known as ideal solutions. The
    solutions that do not obey Raoult’s law (non-ideal solutions)
    have vapour pressures either higher or lower than that predicted by
    Raoult’s law. If the vapour pressure is higher, then the
    solution is said to exhibit positive deviation, and if it is lower,
    then the solution is said to exhibit negative deviation from Raoult’s
    law.


    Vapour
    pressure of a two-component solution showing posit    ive
    deviation from Raoult’s law


    Vapour pressure of a two-component solution showing negative
    deviation from Raoult’s law
    In the
    case of an ideal solution, the enthalpy of the mixing of the pure
    components for forming the solution is zero.
    ΔsolH = 0
    In the
    case of solutions showing positive deviations, absorption of heat
    takes place.
    ∴ΔsolH = Positive
    In the
    case of solutions showing negative deviations, evolution of heat
    takes place.
    ∴ΔsolH = Negative

  • Qstn #15
    An
    aqueous solution of 2% non-volatile solute exerts a pressure of 1.004
    bar at the normal boiling point of the solvent. What is the molar
    mass of the solute?
    Ans : Here,
    Vapour pressure of the
    solution at normal boiling point (p1) = 1.004 bar
    Vapour pressure of pure
    water at normal boiling point
    Mass of solute, (w2)
    = 2 g
    Mass of solvent
    (water), (w1) = 98 g
    Molar mass of solvent
    (water), (M1) = 18 g mol-1
    According to Raoult’s
    law,

    = 41.35 g mol-1
    Hence, the molar mass
    of the solute is 41.35 g mol-1.
  • Qstn #16
    Heptane
    and octane form an ideal solution. At 373 K,
    the vapour pressures of the two liquid components are 105.2 kPa and
    46.8 kPa respectively. What will be the vapour pressure of a mixture
    of 26.0 g of heptane and 35 g of octane?
    Ans : Vapour pressure of
    heptane
    Vapour pressure of
    octane =
    46.8 kPa
    We know that,
    Molar mass of heptane
    (C7H16) = 7 × 12 + 16 × 1
    = 100 g mol-1
    Number
    of moles of heptane
    = 0.26 mol
    Molar mass of octane
    (C8H18) = 8 × 12 + 18 × 1
    = 114 g mol-1
    Number
    of moles of octane
    = 0.31 mol
    Mole fraction of
    heptane,
    = 0.456
    And, mole fraction of
    octane, x2 = 1 - 0.456
    = 0.544
    Now, partial pressure
    of heptane,
    = 0.456 × 105.2
    = 47.97 kPa
    Partial pressure of
    octane,
    = 0.544 × 46.8
    = 25.46 kPa
    Hence, vapour pressure
    of solution, ptotal = p1 + p2
    = 47.97 + 25.46
    = 73.43 kPa
  • Qstn #17
    The
    vapour pressure of water is 12.3 kPa at 300 K.
    Calculate vapour pressure of 1 molal solution of a non-volatile
    solute in it.
    Ans : 1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
    Molar mass of water = 18 g mol-1
    Number of moles present in 1000 g of water
    = 55.56 mol
    Therefore, mole fraction of the solute in the solution is
    .
    It is given that,
    Vapour pressure of water, = 12.3 kPa
    Applying the relation,

    ⇒ 12.3 - p1 = 0.2177
    ⇒ p1 = 12.0823
    = 12.08 kPa (approximately)
    Hence, the vapour pressure of the solution is 12.08 kPa.
  • Qstn #18
    Calculate
    the mass of a non-volatile solute (molar mass 40 g mol-1)
    which should be dissolved in 114 g octane to reduce its vapour
    pressure to 80%.
    Ans : Let the vapour pressure
    of pure octane be
    Then, the vapour
    pressure of the octane after dissolving the non-volatile solute is
    Molar mass of solute, M2 = 40 g mol-1
    Mass of octane, w1 = 114 g
    Molar mass of octane,
    (C8H18), M1 = 8 × 12 +
    18 × 1
    = 114 g mol-1
    Applying the relation,

    Hence, the required
    mass of the solute is 8 g.
  • Qstn #19
    A solution
    containing 30 g of non-volatile solute exactly in 90 g of water has a
    vapour
    pressure of 2.8 kPa at 298 K.
    Further, 18 g of water is then added to
    the
    solution and the new vapour pressure becomes 2.9 kPa at 298 K.
    Calculate:
    1. molar
      mass of the solute
    2. vapour
      pressure of water at 298 K.
    Ans : (i) Let,
    the molar mass of the solute be M g mol-1
    Now,
    the no. of moles of solvent (water),
    And,
    the no. of moles of solute,

    Applying the relation:

    After
    the addition of 18 g of water:

    Again, applying the relation:

    Dividing
    equation (i) by (ii),
    we have:

    Therefore,
    the molar mass of the solute is 23 g mol-1.
    (ii) Putting
    the value of ‘M’ in equation (i),
    we have:

    Hence, the vapour pressure of water at 298 K is 3.53 kPa.
  • Qstn #20
    A
    5% solution (by mass) of cane sugar in water has freezing point of
    271 K.
    Calculate the freezing point of 5% glucose in water if freezing point
    of pure water is 273.15 K.
    Ans : Here, ΔTf = (273.15 - 271) K
    = 2.15 K
    Molar mass of sugar
    (C12H22O11) = 12 × 12 + 22 ×
    1 + 11 × 16
    = 342 g mol-1
    5% solution (by mass)
    of cane sugar in water means 5 g of cane sugar is present in (100 -
    5)g = 95 g of water.
    Now, number of moles of
    cane sugar
    = 0.0146 mol
    Therefore, molality of
    the solution,
    = 0.1537 mol kg-1
    Applying the relation,
    ΔTf = Kf × m

    = 13.99 K kg
    mol-1
    Molar of glucose
    (C6H12O6) = 6 × 12 + 12 ×
    1 + 6 × 16
    = 180 g mol-1
    5% glucose in water
    means 5 g of glucose is present in (100 - 5) g = 95 g of water.
    Number of moles of glucose
    = 0.0278 mol
    Therefore, molality of
    the solution,
    = 0.2926 mol kg-1
    Applying the relation,
    ΔTf = Kf × m
    = 13.99 K kg mol-1 × 0.2926 mol kg-1
    = 4.09 K
    (approximately)
    Hence, the freezing
    point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.
  • Qstn #21
    Two
    elements A and B form compounds having formula AB2 and AB4.
    When dissolved in 20 g of benzene (C6H6),
    1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4 lowers it by 1.3 K.
    The molar depression constant for benzene is 5.1 Kkg mol-1.
    Calculate atomic masses of A and B.
    Ans : We know that,

    Then,
    = 110.87 g mol-1

    = 196.15 g mol-1
    Now, we have the molar
    masses of AB2 and AB4 as 110.87 g mol-1 and 196.15 g mol-1 respectively.
    Let
    the atomic masses of A and B be x and y respectively.
    Now,
    we can write:

    Subtracting equation
    (i) from (ii), we have
    2y = 85.28
    ⇒ y =
    42.64
    Putting the value of
    ‘y’ in equation (1), we have
    x + 2 ×
    42.64 = 110.87
    ⇒ x =
    25.59
    Hence, the atomic
    masses of A and B are 25.59 u and 42.64 u respectively.
    Page No 61:
  • Qstn #22
    At 300 K, 36 g of
    glucose present in a litre of its solution has an osmotic pressure of
    4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the
    same temperature, what would be its concentration?
    Ans : Here,
    T = 300 K
    π = 1.52 bar
    R = 0.083 bar L K-1 mol-1
    Applying the relation,
    π = CRT

    = 0.061 mol
    Since the volume of the
    solution is 1 L, the concentration of the solution would be 0.061 M.
  • Qstn #23
    Suggest the most
    important type of intermolecular attractive interaction in the
    following pairs.
    (i) n-hexane and n-octane
    (ii) I2 and CCl4
    (iii) NaClO4 and water
    (iv) methanol and acetone
    (v) acetonitrile (CH3CN)
    and acetone (C3H6O).
    Ans : (i) Van der Wall’s forces of attraction.
    (ii) Van der Wall’s forces of attraction.
    (iii) Ion-diople interaction.
    (iv) Dipole-dipole interaction.
    (v) Dipole-dipole interaction.
  • Qstn #24
    Based
    on solute-solvent interactions, arrange the following in order of
    increasing solubility in n-octane and explain. Cyclohexane, KCl,
    CH3OH,
    CH3CN.
    Ans : n-octane is a
    non-polar solvent. Therefore, the solubility of a non-polar solute is
    more than that of a polar solute in the n-octane.
    The order of increasing
    polarity is:
    Cyclohexane < CH3CN
    < CH3OH < KCl
    Therefore, the order of
    increasing solubility is:
    KCl < CH3OH
    < CH3CN < Cyclohexane
  • Qstn #25
    Amongst the following compounds, identify which are insoluble, partially
    soluble and highly soluble in water?
    (i) phenol (ii) toluene (iii) formic acid
    (iv) ethylene
    glycol (v) chloroform (vi) pentanol.
    Ans : (i) Phenol (C6H5OH) has the polar group
    -OH and non-polar group -C6H5.
    Thus, phenol is partially soluble in water.
    (ii) Toluene
    (C6H5-CH3) has no polar
    groups. Thus, toluene is insoluble in water.
    (iii) Formic acid (HCOOH) has the polar group -OH and
    can form H-bond with water. Thus, formic acid is highly soluble in
    water.
    (iv) Ethylene glycol has polar -OH group and can form H-bond. Thus, it is
    highly soluble in water.
    (v) Chloroform
    is insoluble in water.
    (vi) Pentanol (C5H11OH) has polar -OH
    group, but it also contains a very bulky non-polar ­­­-C5H11 group. Thus, pentanol is partially soluble in water.
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