Solutions

  • Question 1
    CBSEENCH12005596

    Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

    Solution
    Mass % of benzene
                          = mass of benzenemass of solution ×100= 2222+122×100= 22144×100 = 15.28%
    Mass% of carbon tetrachloride = 100 - 15.28
                              = 84.72%
    Question 2
    CBSEENCH12005597

    Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

    Solution

    Let the total mass of the solution be 100g and mass of benzene be 30 g
    therefore mass of tetrachloride= (100-30)g = 70g
    Molar mass of benzene,



    Question 3
    CBSEENCH12005598

    Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.

    Solution

    solution;

    Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.


    (a) Mol. mass of Co(NO3). 6H2O

                   =58.9+(14+3×16)2+6(18)=58.9+(14+48)×2+108=58.9+124+108 = 290.9

    Moles of Co(NO)3.6H2O
                                           =30290.9=0.103 mol.
    Volume of solution = 4.3 L
    Molarity, 
              M=Moles of soluteVolume of solution in litre    = 1034.3 = 0.024 M

    (b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
    therefore number of moles present in 30ml of 0.5M H2SO4=0.5×301000mol =0.015mol
    therefore molarity =0.015/0.5L 

    thus molarity is 0.03M

                     
      

    Question 4
    CBSEENCH12005599

    Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.

    Solution
    Solution:

    Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

    = Moles of soluteMass of solvent in kg
     
    Mol. mass of urea NH2CONH2
                     = 14 + 2 + 12 + 16 + 14 + 2
                     = 60 g mol-1

    Molality (m) = Moles of soluteMass of solvent in kg

    25 = Moles of solute2.5

    or Moles of solute
                    = 0.25 x 0.25 =  0.625

      Mass of urea
                       = Moles of solute x Molar mass

                       = 0.625 x 60 = 37.5 g

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