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NEET-XII-Chemistry

02: Solutions

with Solutions -
Qstn# A-5 Prvs-QstnNext-Qstn
  • #5
    Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is
    1.202 g mL-1. (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is
    1.202 g mL-1.
    Ans : (a) Molar
    mass of KI = 39 + 127 = 166 g mol-1
    20%
    (mass/mass) aqueous solution of KI means 20
    g of KI is present in 100 g of solution.
    That is,
    20
    g of KI is present in (100 - 20) g of
    water = 80 g of water
    Therefore,
    molality of the solution

    = 1.506 m
    =
    1.51 m (approximately)
    (b) It
    is given that the density of the solution = 1.202 g mL-1
    ∴Volume
    of 100 g solution

    = 83.19 mL
    =
    83.19 × 10-3 L
    Therefore,
    molarity of the solution
    = 1.45 M
    (c) Moles
    of KI
    Moles
    of water
    Therefore,
    mole fraction of KI

    = 0.0263
    Page No 41: (a) Molar
    mass of KI = 39 + 127 = 166 g mol-1
    20%
    (mass/mass) aqueous solution of KI means 20
    g of KI is present in 100 g of solution.
    That is,
    20
    g of KI is present in (100 - 20) g of
    water = 80 g of water
    Therefore,
    molality of the solution

    = 1.506 m
    =
    1.51 m (approximately)
    (b) It
    is given that the density of the solution = 1.202 g mL-1
    ∴Volume
    of 100 g solution

    = 83.19 mL
    =
    83.19 × 10-3 L
    Therefore,
    molarity of the solution
    = 1.45 M
    (c) Moles
    of KI
    Moles
    of water
    Therefore,
    mole fraction of KI

    = 0.0263
    Page No 41:
  • #5-a
    molality
    Ans : Molar
    mass of KI = 39 + 127 = 166 g mol-1
    20%
    (mass/mass) aqueous solution of KI means 20
    g of KI is present in 100 g of solution.
    That is,
    20
    g of KI is present in (100 - 20) g of
    water = 80 g of water
    Therefore,
    molality of the solution

    = 1.506 m
    =
    1.51 m (approximately)
  • #5-b
    molarity and
    Ans : It
    is given that the density of the solution = 1.202 g mL-1
    ∴Volume
    of 100 g solution

    = 83.19 mL
    =
    83.19 × 10-3 L
    Therefore,
    molarity of the solution
    = 1.45 M
  • #5-c
    mole fraction of KI if the density of 20% (mass/mass) aqueous KI is
    1.202 g mL-1.
    Ans : Moles
    of KI
    Moles
    of water
    Therefore,
    mole fraction of KI

    = 0.0263
    Page No 41: