SOLUTIONS QUIZ 15

 

SOLUTIONS QUIZ 16

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Q1.  The weight of H₂C₂O₄ ∙ 2H₂O required to pressure 500 mL of 0.2 N solution is

•   63 g
•   6.3 g
•   0.63 g
•   126 g

Solution
Eq. wt. of H₂C₂O₄ ∙ 2H₂O 

 = 2 + 24 + 64 + 2(2 + 16) /2 = 63  

 𝑤 = 𝑁𝐸𝑉 / 1000 = (0.2 × 63 × 500)/ 1000 = 6.3 g

Q2. The elevation of boiling point method is used for the determination of molecular weight of:

•   Non-volatile and soluble solute
•   Non-volatile and insoluble solute
•   Volatile and soluble solute
•   Volatile and insoluble solute

Solution
To show colligative properties solute should be non-volatile and soluble in given solvent.

Q3.   50 𝑐𝑚³of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding 50 𝑐𝑚³of NaOH. The remaining titration is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is

•    12 𝑐𝑚³
•   10 𝑐𝑚³
•   25 𝑐𝑚³
•   10.5 𝑐𝑚³

Solution
When 0.1 n NaOH is used, 

       𝑁₁𝑉₁ = 𝑁₁𝑉₂ 

 (For HCl) (For KOH) 

0.2𝑁 × 𝑉₁ = 50 × 0.1𝑁 

 𝑉₁ = 50 ×0.1 /0.2 = 25𝑐𝑚³ 

 When 0.5 N KOH is used, 

                       𝑁₁𝑉₁ = 𝑁₃𝑉₃ 

 (For remaining HCl) (for KOH) 

 0.2𝑁 × 25 = 0.5 𝑁 × 𝑉₃ 𝑉₃ = (0.2 ×25) / 0.5 = 10𝑐𝑚³

Q4.  Dissolution of a solute is an exothermic process if :

•   Hydration energy > lattice energy
•   Hydration energy < lattice energy
•   Hydration energy = lattice energy
•   None of the above

Solution
∆𝐻_solution = ∆_{𝐻hydration} + ∆𝐻_{lattice energy}

 ∆𝐻_h = −ve 

∆𝐻_l = +ve

Q5. Which of the following shows maximum depression in freezing point?

•   𝐾₂𝑆𝑂₄
•   NaCl
•   Urea
•   glucose

Solution
  Depression in freezing point is a colligative property. It depends on number of particles. More the number of particles, more will be depression in freezing point. 

 1. 𝐾₂𝑆𝑂₄ → 2𝐾+ + 𝑆𝑂₄ 2− It gives 3 particles. 

2. 𝑁𝑎𝐶𝑙 → 𝑁𝑎+ + 𝐶𝑙− It gives 2 particles. 

 3. Urea→ No dissociation 4. Glucose → No dissociation. 

∴ 𝐾₂𝑆𝑂₄ produces maximum number of particles 

∴ 𝐾₂𝑆𝑂₄ has maximum depression in freezing point. .

Q6.  Water will boil at 101.5 ͦC at which of the following pressure?

•   76 cm of Hg
•   76 mm of Hg
• > 76 cm of Hg
•   < 76 cm of Hg

Solution
Water boils at higher temperature than its b. p. if atmosphere pressure is more than 1 atm.

Q7. The vapour pressure of pure liquid A is 0.80 atm. When a non-volatile B is added to A its vapour pressure drops to 0.60 atm. The mole fraction of B in the solution is  

 

•   0.125
•   0.25
•   0.5
•   0.75

Solution

According to the Raoult’s law the relative lowering vapour pressure which is produced by dissolving a non-volatile solute in a solvent is equal to mole fraction of the solute. 

 𝑃−𝑃_s/𝑃 = 𝑋𝐵 

 where, p= vapour pressure of solvent 

 𝑃/𝑠 = vapour pressure of solution 

𝑋𝐵 = mole fraction of 𝐵 

Given, P=0.80 atm 𝑃_s = 0.60 atm 

 ∴ 𝑋_B = 0.80−0.60/0.80

 = 0.20/0.80 

 = 0.25

Q8. One gram of silver gets distributed between 10 cm3 of molten zinc and 100cm3of molten lead at 8000℃. The percentage of silver still left in the lead layer in approximately NaOH solution

•   2
• 5
•   3
•   1

Solution
𝐾𝐷 = 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑖𝑛 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 / 𝐴 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑖𝑛 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝐵 

 Concentration of Ag in 10 𝑐𝑚₃ Zn=x 

Concentration of Ag in 100 𝑐𝑚₃ Pb=1−𝑥 /10 

Concentration of Ag in 10 𝑐𝑚₃ Pb=1−𝑥 / 10

 300= 𝑥×10 /(1−𝑥) 𝑜𝑟

 𝑥 = 300/ 100 = 0.967 

 = 97% 

Concentration of Ag in zinc =1 -0.967=0.033 =3.3%

Q9. How many moles of 𝐴𝑙₂ (𝑆𝑂₄ )₃ would be in 50 g of the substance?

•   0.083 mol
•   0.952 mol
•   0.481 mol
•   0.140 mol

Solution
Moles = mass molecular weight 

 Given, mass of 𝐴𝑙2 (𝑆𝑂₄)₃ = 50 g 

molecular mass of 𝐴𝑙₃ (𝑆𝑂₄)₃ = 342 

∴ Moles of 𝐴𝑙₂ (𝑆𝑂₄)₃ = 50 /342 = 0.14 mol

Q10.  The distribution law was given by :  

 

•   Henry
•   Nernst
•   van’t Hoff
•   Ostwald

Solution
It is therefore also known as Nernst distribution law