SOLUTIONS QUIZ-10

Q1. 3.65 g of HCl is dissolved in 16.2 g of water. The mole fraction of HCl in the resulting solution is

•  0.4
•  0.3
•  0.2
•  0.1

Solution
X=n/(n+N) n=w/m=3.65/36.5=0.1 N=W/M=16.2/18=0.9 X=0.1/(0.1+0.9)=0.1

Q2.If molecular interaction of two different liquid molecules are stronger than the molecular interactions of the same liquid molecules the mixture is expected to show :

•  Positive deviations
•  Negative deviations
•  No deviations
•  Positive as well as negative deviations

Solution
The tendency to evaporation will decrease and this will lead to lower value of experimental vapour pressure than those calculated from Raoult’s law.

Q3.  Equimolal solutions will have the same boiling point, provided they do not show :

•   Electrolysis
•  Association
•  Dissociation
•  Association or dissociation

Solution
The formula ∆T=K_b ×molality is valid when solute neither dissociates nor associate. In case of dissociation : ∆T=K_b ×molality (1 ⎯ 𝛂 + 𝑥𝛂 + y𝛂). In case of association ∆T=K_b ×molality (1 ⎯ 𝛂 + 𝛂 /n). In case of association

Q4. 6 g urea is dissolved in 90 g water. The relative lowering of vapour pressure is equal to :

•  0.0202
•  1.10
•  0.06
•  0.0196

Solution
(P_(0 )-P_s)/P_0 =(6/60)/(6/60 +90/18)=1/51=0.0196

Q5.At 88 ͦC benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will boil at 88 ͦC at 1 atm pressure, benzene-toluene form an ideal solution?

•  0.740
•  0.688
•  0.588
•  0.416

Solution
 P_m=760 torr, because solution boils at 88 ͦC. Now, 760=900 ×m.f.of C_6 H_6+360 ×(1-m.f.of C_6 H_6) ∴ 760=900a+360-360a; ∴ a=0.74 Where a is mole fraction of C_6 H_6.

Q6. Which of the following aqueous solutions produce the same osmotic pressure? (i)0.1 M NaCl solution (ii) 0.1 M glucose solution (iii)0.6 g urea in 100 mL solution (iv)1.0 g of a non-electrolyte solute (X) in 50 mL solution (molar mass of X=200)

•  (i), (ii), (iii)
•  (ii), (iii), (iv)
• (i), (ii), (iv)
•  (i), (iii), (iv)

Solution
(ii) 0.1 M glucose, π=CRT=0.1RT (iii) 0.6 g urea in 100 mL solution π=n/V RT=(w/m)/V RT=(0.6/60×1000)/100×RT=0.1RT (iv)1.0 g of non electrolyte solute (x) is 50 mL solution π=(1.0/200)/50×1000RT=0.1RT Hence, option (ii), (iii), (iv) have some osmotic pressure, osmotic pressure of 0.1 M NaCl is higher than (ii), (iii), (iv) because it dissociates to give maximum number

Q7.When 20 g of naphthoic acid (C_11 H_8 O_2) is dissolved in 50 g of benzene (k_f=1.72 K kg mol^(-1)), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is

•  3
•  1
•  0.5
•  2

Solution
Actual molecular weight of naphthoic acid (C_11 H_8 O_2)=172 Molecular mass (calculated) =(1000×k_f×w)/(W×∆T_f ) =(1000 ×1.72× 20)/(50×2)=344 van’t Hoff factor (i) =(actual mol.wt.)/(calculated mo.wt.) =172/344 =0.5

Q8.The depression in f.p. is directly proportional to :

•  Normality
•  Molality
•  Molarity
•  None of these

Solution
∆T_b ∝molality

Q9.Which of the following solutions has the highest normality?

•  8 g of KOH/L
•  N phosphoric acid
•  0.5 M H_2 SO_4
•  6 g of NaOH/100 mL

Solution
N=(6×1000)/(40×100)=1.5 N

Q10. The vapour pressure will be lowest of

•  Hypertonic solution
•  Isotonic solution
•  Hypotonic solution
• None of the above

Solution
Living cells shrinks in hypertonic solution (plasmolysis) while bursts in hypotonic solutions (endosmosis). There is no effect when living cells are kept in isotonic solution.