SOLUTIONS QUIZ 18
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Q1. Mole fraction of solute in an aqueous solution which boils at 100.104. 𝐾b for H2O = 0.52 K molality-1:
Solution
∆𝑇 = 𝐾𝑏 × 1000 × 𝑤 × 18/𝑚 × 𝑊 × 18
∆𝑇 = 𝐾𝑏 × 1000 × 𝑤 × 18/𝑚 × 𝑊 × 18
∴ 0.104 =0.52 ×1000 × 𝑛 /𝑁 ×18
∴ 𝑛/𝑁= 3.6 × 10-3
or 1 +𝑛/𝑁 = 1.0036
or
𝑁/𝑛 + 𝑁= 0.996
∴ 𝑛/𝑛+𝑁 = 0.004
Q2. Colligative properties of a solution depends upon
Solution
The properties of solution which depend only on the number of solute particles but not on the nature of the solute taken are called colligative properties.
The properties of solution which depend only on the number of solute particles but not on the nature of the solute taken are called colligative properties.
Q3. X is dissolved in water. Maximum boiling point is observed when X is ….(0.1 M each)
Solution
Elevation in boiling point is colligative property and depends upon number of ions of molecules or particles.
Elevation in boiling point is colligative property and depends upon number of ions of molecules or particles.
𝐶𝑎𝑆𝑂4 → 𝐶𝑎2+ + 𝑆𝑂4 2- ∴ 2 ions
𝐵𝑎𝐶𝑙2 → 𝐵𝑎2+ + 2𝐶𝑙− ∴ 3 ions
NaCl → 𝑁𝑎+ + 𝐶𝑙- ∴ 2 ions
urea → no dissociation ∴ 1 molecule
∵ 𝐵𝑎𝐶𝑙2furnishes maximum ions.
∴ 𝐵𝑎𝐶𝑙2 will have maximum boiling point.
Q4. Lowering of vapour pressure is highest for
Solution
𝑝0 − 𝑝s /𝑝0 = molality × (1 − 𝛼 + 𝑥𝛼 + 𝑦𝛼)
𝑝0 − 𝑝s /𝑝0 = molality × (1 − 𝛼 + 𝑥𝛼 + 𝑦𝛼)
The value of 𝑝0 − 𝑝s is maximum for BaCl2
Q5. The statement “the relative lowering of the vapour pressure is equal to th ratio ot moles of the solute to the total number of the moles in the solution” refers to
Solution
According to Raoult’ law, 𝑝−𝑝s / 𝑝 = 𝑛/𝑛+𝑁
According to Raoult’ law, 𝑝−𝑝s / 𝑝 = 𝑛/𝑛+𝑁
Q6. A solution of two liquids boils at a temperature more than the boiling point of either them. Hence, the binary solution shows
Solution
Negative deviation from Raoult’s law
Negative deviation from Raoult’s law
Q7. Which of the following is true when components forming an ideal solution are mixed?
Solution
∆𝐻m = ∆𝑉m = 0
∆𝐻m = ∆𝑉m = 0
Q8. The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is
NaOH solution
Solution
Osmotic pressure
Osmotic pressure
Q9. The normality of 10% (weight/volume) acetic acid is
Solution
𝑁 = 𝑤 × 1000 / eq.wt.×𝑉(mL) = 10 × 1000 / 60 × 100 = 1.66 N
𝑁 = 𝑤 × 1000 / eq.wt.×𝑉(mL) = 10 × 1000 / 60 × 100 = 1.66 N
Q10. Two solutions of substance (non-electrolyte) are mixed in the following manner. 480 mL of 1.5 M first solution +520 mL of 1.2 M second solution. What is the molarity of the final mixture ?
Solution
Total molarity = 𝑀1𝑉1+ 𝑀2𝑉2 / 𝑉1+ 𝑉2
Total molarity = 𝑀1𝑉1+ 𝑀2𝑉2 / 𝑉1+ 𝑉2
=
1.5 × 480+1.2 × 520 /480+520
= 1.344 m