SOLUTIONS [ NCERT INTEXT EXERCISES ]

                YASH CLASSES

                   SOLUTIONS     

     NCERT INTEXT EXERCISES

  Q1.Calculate the mass percentage of benzene

       and carbon tetrachloride if 22g of benzene 

       is dissolved in 122g of carbon tetrachloride.

  Ans-  Mass of solution =Mass of benzene + Mass of CCl₄

                                       = 22g + 122g  = 144g

          (1)Mass of benzene Mass of benzene 100 Mass of

                 solution = (22/144) x100 = 15.28% .

          (2)Mass % of  CCl₄  =100 - 15.28 = 84.72%

 

Q.2. Calculate the mole fraction of benzene in solution

       containing 30% by mass in carbon tetrachloride.

 

   Ans-  Let A - Benzene:    B – CCI₄

     W{A} = 30g M{A} = 78,  n{A} = W{A}/M{A} = 30/78 = 0.38    

    W{B} = 70g M{B} = 154 , n{B} = W{B}/M{B} = 70/154 = 0.45   

    X{A} = n{A}/(n{A} + n{B}) = 0.38/(0.38 + 0.45) = 46

 [ W{A},W{B} given mass , M{A},M{B}molar mass, 

       n{A},n{B}moles   ,  X{A} mole fraction]

Q.3. Calculate the molarity of each of the following solutions:

    (a) 30 g of Co (NO₃).6H₂O in 4.3 L of solution,

    (b) 30 ml of 0.5M H₂SO₄ diluted to 500 ml.

  Ans .(a) Molar mass of Co (NO₃).6H₂O = M_B = 291

            W_B of Co (NO₃).6H₂O = 30g, V_sol = 4.3L : 4300ml

           M = (W_B x1000) /( M_B X V_Sol )

                = (30 X 1000)/(291 X 4300) = 0.024

          (b) Molarity of solution after dilution may be calculated as

                 M₁ V₁ Concentrated) = M₂ V₂ (Diluted)

                                      0.5 X30 = M₂X 5

                                 M₂ = (0.5 X 30)/500 = 0.03 moles /litre 

  Q. 4. Calculate the mass of urea ( NH₂CONH ₂)= required in

          making 2.5 kg of 0.25 molal aqueous solution.

  Ans- 0.25 molal solution means 0.25 moles of urea in 1000g of water:   

             Mass of urea 0.25 x 60 = 15g

    Total mass of solution I = 1000 + 15 = 1015g = 1.015kg

       1.015 kg of solution contain urea = 15g

       2.5 kg of solution will require urea = (15x2.5)/1.015 = 36.94g.

Q.5. Calculate (a) molality (b) molarity and (c) mole fraction of

       KI if the density of 20% (mass/mass) aqueous Kl is 1.202g ml^-1.

  Ans- Here.  M_B =166u , W_B  = 20g , W_A = 80g

           (a) m = (W_B x1000) /( M_B X W_A)

            m= (20 x1000) /( 166 X 80 )  , m = 1.5mol^-1.

         (b) M = (% x density x10) / M_B  ,

              M = (20 x1.202x 10) / 166  ,  M= 1.45mol L^-1.

       (c)  moles of KI =  given mass of KI / molar mass of KI

             Moles of KI = 20 /166 = 0.120 ( given 20% KI by mass)

              Moles of water = 80 /18 = 4.44 ( so, 80% water by mass)

        moles fraction of KI = moles of KI / ( moles of KI + moles of water)

                                     X_KI = 0.120 / (4.44 + 0.120) = 0.0263

 Q.6. H₂S a toxic egg like smell is used for the qualitative analysis. 

        If the solubility of H₂S in water at STP is 0.195m, 

         calculate Henry’s law constant.

 Ans-  0.195m solution means that 0.195 moles of H₂S is

          dissolved in 1 kg of water.  Moles of H₂S = 0.195

            Moles of water  = 1000/18 = 55.5S

     

    Q.7. Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 295 K.  

           Calculate the quantity of CO₂ in 500 ml of soda water when

            packed under 2.5 atm CO₂ pressure at 298 К.

 

     Ans- According to Henry's law P = K_H X  - ①

            given  P = 2.5 atm = 2.5 x101325Pa ,  K_H =  1.67 x 10⁸ Pa

             Putting these values in equation ①

            X = P/ K_H =  2.5 x101325Pa/1.67 x 10⁸ Pa

              mole fraction of CO₂ =  1.517 x 10^-3 .

           moles of H₂O = 500/ 18 = 27.77

    mole fraction of CO₂ = moles of CO₂ / (moles of CO₂ + moles of  H₂O)

                       moles of H₂O >>> moles of CO₂

           so,  mole fraction of CO₂ = moles of CO₂ / moles of  H₂O

                   1.517 x 10^-3 = moles of CO₂ / 27.77

                  moles of CO₂ =  1.517 x 10^-3 x 27.77 = 0.042 mol.

 Q 8. The vapour pressure of pure liquids A and Bare  450 mm and 700 mm  

       Hg respectively 350 K. Find out the composition of the liquid mixture

        if total vapour pressure is 600 mm Hg. Also find the composition of   

       vapour phase.

    Ans- Here P⁰_A = 450 mm ,  P⁰_B = 700 mm , P _Total = 600 mm

          Applying Raoult’s law  ,    P_A = X_A P⁰_A , ① 

                   X_A + X_B =1 ② and    P_B = X_B P⁰_B ③

                       from equations② and③    P_B = (1- X_A)P⁰_B ④

             P _Total = P_A + P_B , put   P_A and  P_B values by equations ① and④

                P _Total = X_A P⁰_A +(1- X_A)P⁰_B ,

                 So  P _Total = P⁰_B +( P⁰_A - P⁰_B) X_A ,

                     600 = 700 + (450 -700)  X_A ,

                    X_A = 0.40  

         therefore mole fraction of  X_B= 1- 0.40 =0.60 ( by equation ②)

           P_A = X_A P⁰_A ,   P_A = 0.40 x 450 = 420 mm

            mole fraction of A in vapour phase =   P_A / ( P_A +P_B ),

                                           = 180 / (180 + 420) = 0.30

           mole fraction of B in vapour phase = 1- 0.30 = 0.70

   Q. 9. Vapour pressure of pure water at 298K is  23.8 mm Hg. 50 g of 

           urea (NH₂CONH ₂) dissolved in 850 g of water, Calculate 

        vapour pressure of water for this solution and its relative lowering.

 Ans – We know lowering vapour pressure ,

        (P⁰ – P_s) / P = n₂ /(n₁ + n₂ ) = (W₂ / M₂ )/[(W₁/M₁ ) + (W₂ / M₂ )]

        P⁰ = 23.8 mm , W₂ = 50g , M₂ (urea) = 60g mol^-1 , W₁= 850 g

          M₁ (water) =  850 g mol^-1 , [ where n₁ , n₂ are moles]

          for calculate P_s , (P⁰ – P_s) / P = (50/60)/ [(850/18) + (50/60)]

                                                         = 0.017

  Q10. boiling point of water  750 mm Hg is 99.63⁰ C. how much sucrose

       is to be added to 500g of water such that it boils at 100⁰C. Molal

       elevation constant for water is 0.52 K kg mol^-1 .

     Ans - ∆T_b = 100 – 99.63 = 0.37 , K _b = 0.52 K kg mol^-1 .

               W_B = ? , M_B (mass of sucrose C₁₂ H₂₂ O₁₁ ) = 342 g mol^-1 ,

                     W_A (given mass of solvent ,water ) = 500 g

             Amount of required sucrose may  - ∆T_b = K _b x m (molality)

              be added   ∆T_b = ( K _b x W_B x 1000) / ( M_B x W_A )

                               0.37 = ( 0.52 x W_B x 1000) / (342 x 500)

                                 W_B = 121.67g.

   Q. 11. Calculate the mass of ascorbic acid ( vitamin C, C₆H₈O₆) to be

          added in 75g of acetic acid to lower its melting point by 1.5⁰ C.

           K _f = 3.9 K kg mol^-1 .

          Ans -∆T_f = 1.5⁰ C.  , K _f = 3.9 K kg mol^-1 .

               W_B = ? , M_B (mass of ascorbic acid  C₆H₈O₆) = 176 g mol^-1 ,

                     W_A (given mass of solvent ,water ) = 75 g

    Amount of required sucrose may  - ∆T_f = K _f x m (molality)

              be added    ∆T_b = ( K _b x W_B x 1000) / ( M_B x W_A )

                               1.5 = ( 3.9 x W_B x 1000) / (176 x 75)

                                 W_B = 5.077g.

  Q.12. Calculate the osmotic pressure in Pascals exerted by a solution

          prepared by dissolving 1.0g of polymer of molar mass 1,85,000

        in 450 ml of water at 37⁰ C.

    Ans-  osmotic pressure  = CRT , C = n / V ,

          so,   = nRT /V ,   given V = 450 ml , T = 37 + 273= 310K,

              R(Rydberg constant) = 8.314 x 10³ Pa L g mol^-1K^-1 ,

               n = 1.0/ 185000

                = (1x 8.314x 10³ x 310) / 185000 x (450/1000)

                n = 30.96 Pa .