Solved Examples - Solutions (Part - 4) | Additional Study Material for JEE PDF Download

Example 45. Sea water is 3.5% by mass of a salt and has a density 1.04 g cm^-3 at 293 K. Assuming the salt to be sodium chloride, calculate the osmotic pressure of sea water. Assume complete ionization of the salt.
Solution. Mass of NaCl = 3.5 g

Actual number of moles of particles of solute in solution=
Volume of solution = 1 litre

Example 46. Molality of a solution in aqueous medium is 0.8. Calculate its mole fraction and the percentage by mass of solute if molar mass of solute is 60.
Solution. We know that,

where, x_B = mole fraction of solute
m_A = molar mass of solvent

x_B = 0.014
Let w_B = x g, w_A = 100 g

Example 47. A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 56.8 cm³ of benzene (density 0.889 g cm^-3). At room temperature, vapour pressure of this solution is 98.8 mm Hg while that of benzene is 100 mm Hg. Find the molality of the solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene?
Solution. 

ΔT_f = K_f × Molality
0.73 = K_f × 0.1557
K_f = 4.688

Example 48. x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05 M NaCl solution. The osmotic pressure of this solution is found to be 4.92 atm at 27ºC. Calculate the value of `x'. Assume complete dissociation of NaCl and ideal behaviour of this solution.
(a) 16.52 gm      
(b) 24.032 gm
(c) 19.959 gm    
(d) 12.35 gm
Ans. (c)
Solution. 
(i) For NaCl : p = iCRT = 2 × 0.05 × 0.0821 × 300 = 2.463 atm
(ii) For unknown compound,

Total osmotic pressure p = p₁ + p₂
4.92 = 2.463 + 0.1231 x
x = 19.959 g

Example 49. The freezing point of a solution containing 50 cm³ of ethylene glycol in 50 g of water is found to be -34ºC. Assuming ideal behaviour, calculate the density of ethylene glycol (K_f for water = 1.86 K kg mol^-1).
(a) 1.13 g/cm³
(b) 2.00 g/cm³
(c) 1.8 g/cm³
(d) 2.25 g/cm³
Ans. (a)
Solution. 

Example 50. Match the boiling point with K_b for x, y and z if molecular weight of x, y and z are same.

Solution. Molal elevation constant may be calculated as,

 
(where, Tº_b = boiling point of pure solvent L_v = latent heat of vaporization.)

(here, ΔH_V = molar latent heat of vaporization).

here, ΔS_V = entropy of vaprization.
By considering ΔS_V as almost constant, K_b µ Tº.
K_b(x) = 0.68 ; K_b (y) = 0.53 and K_b (z) = 0.98.

Example 51. 1.22 g C₆H₅COOH is added into two solvents and data of ΔT_b and K_b are given as -
(a) ln 100 g CH₃COCH₃; ΔT_b= 0.17; K_b = 1.7 kg kelvin/mol
(b) ln 100 g benzene; ΔT_b = 0.13; K_b = 2.6 kg kelvin/mol
Find out the molecular weight of C₆H₅COOH in both cases and inerpret the result.
Solution. 
(a) 

(b) 

(Abnormally double molecular mass of benzoic acid, it shows association of benzoic acid in benzene).

Example 52. How much C₂H₅OH should be added to 1 litre H₂O so that it will not freeze at -20ºC?
K_f = 1.86ºC/m
Solution. Mass of 1 litre water = 1000 g

Example 53. Calculate the molarity of each of the ions in solution when 3.0 litre of 4.0 M NaCl and 4.0 litre of 2.0 M CoCl₂ are mixed and diluted to 10 litre.
Solution. Molarity Na  = 1.2 M
Molarity Co²⁺  = 0.8 M
Molarity Cl^- = 2.8 M

Total Cl^- ions = 28 mole.

Example 54. Calculate the molarity of each ion in solution after 2.0 litre of 3.0 M AgNO₃ is mixed with 3.0 litre of 1.0 M BaCl₂.
Solution. 
BaCl₂₊  2AgNO₃ → 2AgCl +Ba(NO₃)₂
Initial 3 6 -_
mole
Final - -6 (ppt) 3
mole

Example 55. 1.2 kg ethylene glycol  was added in a car radiator containing 9 litre water. The freezing of water was just prevented when car was running in the Himalayan valley at temperature -4ºC. Sudden thunderstorm in the valley lowered the temperature to -6ºC. Calculate the amount of ice separated.
Solution. 

A → Solute; B → Solvent

w_B = 6000 g
weight of ice = (Total weight of H₂O) - (wt. of H₂O at 6°C) = 9000 - 6000 = 3000 g = 3 kg

Example 56. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be 20 mm of mercury.
(a) 0.8
(b) 0.6
(c) 0.4
(d) 0.2
Ans. (b)
Solution. 
Mole fraction of solute 

Comparing under the two conditions,

or mole fraction of solute = 0.4
mole fraction of solvent = (1 -0.4) = 0.6

Example 57. Three solutions of HCl having normality 12 N, 6 N and 2 N are mixed to obtain a solutions of 4 N normality. Which among the following volume ratio is correct for the above three components?
(a) 1 : 1 : 5
(b) 1 : 2 : 6
(c) 2 : 1 : 9
(d) 1 : 2 : 4
Ans. (b)
Solution. 
Use Hit & Trial Method.
N₁V₁ + N₂V₂ + N₃V₃ = N_R(V₁ + V₂ + V₃)
12 × 1 6 × 2 2 × 6 = N_R(9)
N_R = 4

Example 58. Two solutions of H₂SO₄ of molarities x and y are mixed in the ratio of V₁ mL : V₂ mL to form a solution of molarity M₁. If they are mixed in the ratio of V₂mL : V₁mL, they form a solution of molarity M₂. Given V₁. Given V₁/V₂ => 1 and  = , then x : y is -
(a) 2 : 1
(b) 4 : 1
(c) 1 : 2
(d) 3 : 1
Ans. (a)
Solution.
Molarity of the mixture can be calculated as.
M₁V₁ + M₂V₂ = M_R(V₁ + V₂)
where, M_R = resultant solution
(V₁ × x) × (V₂ × y) = M₁(V₁ + V₂)
(V₂ × x) × (V₁ × y) = M₂(V₁ + V₂)
Dividing equation (i) by equation (ii), we get

Substituting we can calculate x : y.

Example 59. Isuling (C₂H₁₀O₅)_n is dissolved in a suitable solvent and the osmotic pressure (p) of solutions of various concentrations (g/cc) C is measured at 20ºc. The slope of the plot of p against 'C' is found to be 4.65 × 10^-3. The molecular weight of insulin is -
(a) 4.8 × 10⁵
(b) 9 × 10⁵
(c) 3 × 10⁵
(d) 5.17 × 10⁶
Ans. (d)
Solution.

where C = concentration in g/cc,
Comparing eqs. (i) and (ii), ...(ii)

Example 60. Compound PdCl₄.6H₂O is a hydrated complex; 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex (K_f for water = 1.86 K kg mol^-1)
(a) [Pd(H₂O)₆]Cl₄
(b) [Pd(H₂O)₄Cl₂]Cl ₂.2H₂O
(c) [Pd(H₂O)₃Cl₃]Cl.3H ₂O
(d) [Pd(H₂O)₂Cl₄].4H ₂O
Ans. (c)
Solution.
ΔT = i × K_f × m
(273 -269.28) = i × 1.86 × 1
3.72 = i × 1.86
i = 2

Thus, the complex should give two ions in the solution, i.e., the complex will be [Pd(H₂O)₃Cl₃]Cl.3H₂O]

Example 61. pH of 0.1 M monobasic acid is measured to be 2. Its osmotic pressure at a given temperature T K is -
(a) 0.1 RT
(b) 0.11 RT
(c) 1.1 RT
(d) 0.01 RT
Ans. (b)
Solution.
HA  H⁺  A^-
t = 0  C  0     0
t_eq C -C∝ C∝ C∝
[H ] = Ca, [H ] = 10^-pH
Cα = 10^-2
0.1 α = 10^-2
a = 0.1
a =  ; 0.1 =
i = 1.1
p = iCRT
= 1.1 × 0.1 × RT = 0.11 RT

Example 62. Lowering of vapour pressure in 1 molal aqueous solution at 100ºC is -
(a) 13.44 mm Hg
(b) 14.12 mm Hg
(c) 31.2 mm Hg
(d) 35.2 mm Hg
Ans. (a)
Solution.
Molality and mole fraction are related as follows:

m_A = molar mass of solvent
x_B = 0.0176, x_A = 0.9824
p = p°_Ax_A
p = 760 × 0.9824 = 746.624
Δp = p°_A - p = 760 - 746.624 = 13.4 mm Hg.