chemistry

EAMCET PROBLEMS WITH SOLUTIONS FROM SOLUTIONS CHAPTER

Example:          50 g of a saturated aqueous solution of potassium chloride at 30°C is evaporated to dryness, when 13.2 g of dry KCl was obtained. Calculate the solubility of KCl in water at 30°C.

Solution:           Mass of water in solution = (50 – 13.2) = 36.8 g

Solubility of KCl = Mass of KCl/Mass of water × 100 = 13.2/36.8 × 100

Example:          How much copper sulphate will be required to saturate 100 g of a dilute aqueous solution of CuSO₄ at 25°C if 10 g of a dilute solution leave on evaporation and drying 1.2g of anhydrous CuSO₄? The solubility of CuSO₄ in water at 25°C is 25.

Solution:           100 g of dilute solution of CuSO₄ contain

                        = 1.2 × 10 = 12.0 g CuSO₄

Mass of water present in dilute solution

                        = (100 – 12) = 88 g

To saturate 100 g of water, CuSO₄ required = 25 g

So, to saturate 88 g of water, CuSO₄ required 25/100 × 88

                               

                      = 22 g

Thus, the mass of CuSO₄ to be added to 100 g of dilute solution to saturate it = (22 – 12) = 10 g

 Example:

The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure for solution and the mole fraction of methanol in the vapour.

Solution:

Mol. mass of ethyl alcohol = C₂H₂OH = 46

No. of moles of ethyl alcohol = 60/46 = 1.304

Mol. mass of methyl alcohol = CH₃OH = 32

No. of moles of methyl alcohol = 40/32 = 1.25

‘X_A’, mole fraction of ethyl alcohol = 1.304/1.304+1.25 = 0.5107

‘X_B’, mole fraction of methyl alcohol = 1.25/1.304+1.25

Partial pressure of ethyl alcohol = X_A. p_A⁰ = 0.5107 × 44.5

                                           = 22.73 mm Hg

Partial pressure of methyl alcohol = XB.p_A⁰ =0.4893 × 88.7

                                             = 43.73 m Hg

Total vapour pressure of solution = 22.73 + 43.40

                                             = 66.13 mm Hg

Mole fraction of methyl alcohol in the vapour

            = Partial pressure of CH₃OH/Total vapour pressure = 43.40/66.13 = 0.6563

Example :

Two liquids A and B form ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.

Solution:

Let the vapour pressure of pure A be = p_A⁰; and the vapour pressure of pure B be = p_B⁰.

Total vapour pressure of solution (1 mole fraction of A and X_B is mole fraction of B)

550 = 1/4 p_A⁰ + 3/4 p_B⁰

or 2200 = p_A⁰ + 3p_B⁰              …(i)

Total vapour pressure of solution (1 mole A + 3 moles B)

         = 1/5 p_A⁰ + 4/5 p_B⁰

   560 = 1/5 p_A⁰ + 4/5 p_B⁰

  

or 2800 = p_A⁰ + 4p_B⁰                                 …(ii)

Solving Eqns. (i) and (ii),

          p_B⁰ = 600 mm of Hg = vapour pressure of pure B

          p_A⁰ = 400 mm of Hg = vapour pressure of pure A

  

Example:

An aqueous solution containing 28% by mass of a liquid A (mo. Mass = 140) has a vapour pressure of 160 mm at 37^oC. Find the vapour pressure of the pure liquid A. (The vapour pressure of water at 37^oC is 150 mm).

Solution:

For total miscible liquids,

P_total = Mol. fraction of A × p_A⁰ + Mol. fraction of B × p_B⁰

No. of moles of A = 28/140

Liquid B is water. its mass is (100−28), i.e., 72.

No. of moles of B = 72/18

Total number of moles = 0.2 + 4.0 = 4.2

Given, P_total = 160 mm

         p_B⁰ = 150 mm

So 160 = 0.2/4.2 × p_B⁰ + 4.0/4.2 × 150

        p_A⁰ = 17.15×4.2/0.2 = 360.15 mm

Example:

The vapour pressure of two pressure liquids A and B that forms an ideal solution at 300 and 800 torr respectively at temperature T. A mixture of the vapours of A and B for which the mole fraction A is 0.25 is slowly compressed at temperature T. Calculate:

(a) The composition of the first drop of condensation.

(b) The total pressure of this drop formed.

(c) The composition of the solution whose normal boiling point is T.

(d) The pressure when only the last bubble of vapours remains.

(e) Composition of the last bubble.

Solution:

(a) p_B⁰ = 300 torr p_B⁰ = 800 torr

    y_A =pA/P= p_A⁰x_A/P
 

where yA and yB are mole fraction of A and B in the vapour

 y_B = p_B⁰x_A/P
 

∴ y_A = y_B = p_B⁰x_A/p_B⁰x_B

   0.025/0.75 = 300x_A/800x_B

i.e.,  x_A/x_B = 8/9

i.e.,   x_A = 8/17 = 0.47

                                    Composition of last drop

        x_B = 9/17 = 0.53

(B)    Pressure of last drop:

        p = p_A⁰x_A + p_B⁰x_B 

(c)    At boiling point: p = 760

       

        760^o = 300 x_A + 800 (1−x_A)

        x_A = 0.08,          ∴ x_B = 0.92

(d)    For last drop:      x_A = 0.25, x_B = 0.75

        P = 300 × 0.25 + 800 × 0.75 = 675 torr.

(e)    y_A = p_A/P

        y_A = 300×0.25/675 = 0.111

        y_B = 0.889

Here, y_A and y_B are composition of vapour of last drop.

Example:

Calculate the vapour pressure lowering caused by addition of 50 g of sucrose (molecular mass = 342) to 500 g of water if the vapour pressure of pure water at 25°C is 23.8 mm Hg.

 Solution:

According to Raoult’s law,

   p₀–p_s/p₀ = n/n+N

 or Λ_p n/n+N.p₀

 Given n = 50/342 = 0.416, N = 500/18 = 27.78 and p₀ = 23.8

  Substituting the values in the above equation,

 ¦p = 0.146/0.146+27.78 × 23.8 = 0.124 mm Hg

  

Example:

The vapour pressure of pure benzene at a certain temperature is 640mm Hg. A non-volatile solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular mass of the solid substance?

Solution:

According to Raoult’s law.

   p₀–p_s/p₀ = n/n+N 

 Let m be the molecular mass of the solid substance.

 n = 2.175/m ; N = 39/78 = 0.5

 [molecular mass of benzene = 78]

 Substituting the values in above equation.

640–600/640 = 2.175/m/2.175/m+0.5 = 2.174/2.175+0.5m 

m = 2.175×16 – 2.175/0.5 = 65.25

 Example:

 A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25°C. Further 18 g of water is then added to the solution; the new vapour pressure becomes 22.15 mm Hg of at 25°C. Calculate (a) molecular mass o the solute and (b) vapour pressure of water at 25°C.

Solution:

 Let the vapour pressure of water at 25°C be p₀ and molecular mass of the solute be m.

 Using Raoult’s law in the following form.

 For solution (I), (p₀–21.85)/21.85 = 30×18/90×m          …(i)

 For solution (II), (p₀–22.15)/22.15 = 30×18/108×m        …(ii)

 Dividing Eq. (i) by Eq. (ii),

(p₀–21.85)/21.85 × 22.15/(p₀–22.15) = 108/90 = 6/5

Substituting the value of p₀ in Eq. (i)

   M = 67.9

Example:

 What mass of non-volatile solute (urea) needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 5%. What will be the molality of solution?

 Solution:

 Using  Raoult’s law in the following form,

  p₀–p_s/p_s = wM/Wm

 If p₀ = 100 mm, then p_s = 75 mm

 100–75/75 = w×18/100×60

  w = 111.1

Molality = w×1000/m×W = 111.1×1000/60×100 = 18.52 m

Example: On dissolving 10.8 glucose (m.w. = 180) in 240 g of water, its boiling point increases by 0.13^oC. Calculate the molecular elevation constant of water.

Solution: ?T = 100K'×w/W×m

  

              or K' = ?T×W×m/100×w

  

Given, ?T =0.13^oC, W = 240 g, m = 180 and 2 = 10.8 g

  

K' = 0.13×240×180/100×10.8 = 5.2^o

Example: A solution of 2.5 g of a non-volatile solid in 100 g benzene boiled at 0.42^oC higher than the boiling point of pure benzene. Calculate the molecular mass of the substance. Molal elevation constant of benzene is 2.67 K kg mol⁻¹.

  

Solution: m = 1000K_b×w/W×?T

Given, K_b = 2.67, w = 2.5 g, W = 100 g, ?T = 0.42

  

m = 1000×2.67×2.5/100×0.42 = 158.9

  

The molecular mass of substance is 158.9.

Example: The molal elevation constant for water is 0.56 K kg mol⁻¹. Calculate the boiling point of a solution made bt dissolving 6.0 g of urea (NH₂CONH₂) in 200 g of water.

  

Solution:  ?T = 1000K_b×w/m×W

Given, K_b = 0.56 K kg mol⁻¹, w = 6.0 g, W = 200 g, m=60

?T = 1000×0.56×6.0/200×60 = 0.28^oC

Thus, The boiling point of solution = b.pt. of water + ?T = (100^oC + 0.28^oC) = 100.28^oC

Example: By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^oC. Calculate the molecular mass of the substance. [Molal depression constant for water = 1.863 K kg mol⁻¹]

Solution: m = 1000K_f×w/W×?T

Given, K_f = 1.863 K kg mol⁻¹.

         w = 13.6 g, W = 20 g, ?T = 3.7^oC

               m = 1000×1.863×13.6/20×3.7 = 243.39

Example: On dissolving 0.25 g of a non-volatile substance in 30 mL benzene (density 0.8 g/mL), its freezing point decreases by 0.40^oC. Calculate the molecular mass of non-volatile substance (K_f = 5.12 K kg mol⁻¹).

Solution: Mass of benzene, W = volume × density

                                         = 30 × 0.8 = 24 g

Given, Kf = 5.12 K kg mol⁻¹, w = 0.25 g, ?T = 0.40^oC.

We know that

      m = 1000K_f×w/W×?T

         = 1000×5.12×0.25/24×0.40 = 133.33

  

Example: A solution of 1.25 f of a certain non-volatile substance in 20 g of water freezes at 271.94 K. Calculate the molecular mass of the solute. [K_f = 1.86 K kg mol⁻¹]

Solution: Freezing point of water = 273.0 K

?T = (273–271.94) = 1.06K

We know that m = 1000K_f×w/W×?T

Given K_f = 1.86 K kg mol⁻¹, w = 1.25 g, W = 20 g and ?T = 1.06 K.

         m = 1000×1.86×1.25/20×1.06 = 109.66

Example:

Phenol associated in benzene to a certain extent for a dimer. A solution containing 20 × 10⁻³ kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. Calculate the fraction of the phenol that has dimerised. (K_f of benzene is 5.12^oK kg mol⁻¹)

Solution:

Observed mol. mass

  

          = 1000×Kf×w/W?T

  

          = 1000×5.12×20×10^–3/1×0.69 = 148.4

Normal mol. mass of phenol (C₆H₅OH) = 94

So Normal mol. mass/Observed mol. mass = 94/148.4

          = 1 + (1/n + 1)α = 1 + (1/2 – 1)α

94/148.4 = 1–α/2

or α = 0.733 or 73.3 %

  

Example:

The molal depression of the freezing point in 1000 g of water is 1.86. Calculate what would be the depression of freeing point of of water when (a) 120 g of urea. (b) 117 g of sodium chloride, (c) 488.74 g of BaCl₂.2H₂O have been dissolved in 1000 g of water. It is assumed that sodium chloride and barium chloride are fully ionized.

Solution:

(a) Urea is a non-electrolyte. The number of particles does not change in solution.

     Thus, ?T = 1000×K_f×w/W×m = 1000×1.86×120/1000×60 = 3.72^oC

     No. of moles of urea = 120/60 = 2

(b) No. of moles of NaCl = 117/58.5 = 2

     NaCl is an electrolyte. It ionizes compeletely.

     One molecule of NaCl furnishes 2 ions (one Na⁺ and one Cl⁻).

                       

     Hence, the number of particles will be double as compared to urea solution.

     So ?T_NaCl = 2 × ?T_urea = 2 ×3.72 = 6.44^oC

(c) One molecule of BaCl₂.2H₂O furnishes three ions (one Ba²⁺ and two Cl⁻)

     No. of moles of BaCl₂.H₂O = 488.74/244.37 = 2

     Hence, the number of particles will be three times as compared to the urea solution.

     So ?T_BaCl2 = 3?T_urea = 3 × 3.72 = 11.16^oC

Example:

Arginine vasopressin is a pituitary hormone. It helps to regulate the amount of water in the body by reducing the flow of urine form kidneys. An aqueous solution containing 21.6 mg of vasopressin in 100 mL of solution had an osmotic pressure of 3.70 mm Hg at 25^oC. What is molecular weight of hormone?

Solution:

We know πV = nRT

             πV = w_B/m_B RT

             m_B = w_B×RT/πV                                   …(i)

where w_B = mass of solute (21.6× 10⁻³ g)

         m_B = molar mass of solute

         R = 0.0821 L atm K⁻¹ mol⁻¹

         T = 298 K

               V = 100/1000 = 0.1 L : π = 3.70/760 atm

From (i), m_B = 21.6×10^–3×0.0821×298/(3.70/760)×0.1 = 1085 g/mol

  

Example:

A solution is prepared by dissolving 1.08 g of human serum albumin, a protein from blood plasma, in 50 cm³ of aqueous solution. The solution has an osmotic pressure of 5.85 mm Hg at 298 K

(a) What is molar mass of albumin?

(b) What is height of water column placed in solution?

              d(H₂O) = 1gm cm³

Solution:

(a) Molar mass of albumin cab be calculated using followed relation

     m_B = w_B×RT/πV                                                         …(i)

Given w_B = 1.08; R = 0.0821 litre atm K⁻¹ mol⁻¹

              T = 298 K, π = 5.85/760 atm; V = 50/1000 = 0.05 litre

  

Substituting these values in (i)

         m_B = 1.08×0.0821×298/(5.85/760)×0.05 = 68655 g/mol

(b) π h dg

        5.85/760 × 1010325 = h × 1 × 10^–3 × 9.8

     ∴     h = 7.958 × 10⁻² m = 7.958 cm

Example:

Calculate osmotic pressure of 5% solution of cane sugar (sucrose) at 15^oC.

Solution:

m = mol. mass of sucrose (C₁₂H₂₂O₁₁) = 342

w = 5 g,      V = 100 mL = 0.1 litre

S = 0.082,   T = (15 + 273) = 288 K

Applying the equation PV = w/m ST,

  

          P = 5./342 × 1/0.1 × 0.082 × 288

            = 3.453 atm

Example:

The solution containing 10 g of an organic compound per litre showed an osmotic pressure of 1.16 atmosphere at 0oC. Calculate the molecular mass of the compound (S = 0.0821 litre atm per degree per mol.)

Solution:

Applying the equation

         m = w/PV . ST

Given w = 10 g, P = 1.18 atm, V = 1 litre, S = 0.0821 and T = 273 K.

        m = 10/1.18×1 × 0.0821 × 273 = 189.94

  

Example:

The osmotic pressure of a solution containing 30 g of a substance in 1 litre solution at 20^oC is 3.2 atmosphere. Calculate the value of S. The molecular mass of solute is 228.

Solution:

Applying the equation

        PV = w/m . ST

or     S = m×P×V/w×T

Given m = 228, P = 3.2 atm, V = 1 litre, w = 30 g and

        T = 20 + 273 = 293 K

        S = 228×3.2×1/30×293

           = 0.083 litre atm per degree per mol

  

Example:

What is the volume of solution containing 1 g mole of sugar that will give rise to an osmotic pressure of 1 atmosphere at 0^oC ?

Solution:

Applying the equation PV = nST,

          V = n/P × S × T

Given n = 1, P = 1 atm, S = 0.821 and T = 273 K

 V = 1/1 × 0.0821 × 273 = 22.4 litre