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NEET-XII-Chemistry

02: Solutions

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  • #3
    Calculate
    the molarity of each of the following solutions: (a) 30 g of Co(NO3)2.6H2O in 4 .3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
    Ans : Molarity
    is given by:

    (a) Molar
    mass of Co (NO3)2.6H2O
    = 59 + 2 (14 + 3 × 16) + 6 × 18
    = 291 g mol-1
    ∴Moles
    of Co (NO3)2.6H2O
    = 0.103 mol
    Therefore,
    molarity
    = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
    ∴Number
    of moles present in 30 mL of 0.5 M H2SO4
    = 0.015 mol
    Therefore, molarity
    = 0.03 M
  • #3-a
    30 g of Co(NO3)2.6H2O in 4 .3 L of solution
    Ans : Molar
    mass of Co (NO3)2.6H2O
    = 59 + 2 (14 + 3 × 16) + 6 × 18
    = 291 g mol-1
    ∴Moles
    of Co (NO3)2.6H2O
    = 0.103 mol
    Therefore,
    molarity
    = 0.023 M
  • #3-b
    30 mL of 0.5 M H2SO4 diluted to 500 mL.
    Ans : Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
    ∴Number
    of moles present in 30 mL of 0.5 M H2SO4
    = 0.015 mol
    Therefore, molarity
    = 0.03 M