SOLUTIONS QUIZ 13

 

SOLUTIONS Quiz-13

Dear Readers, The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry.

Q1.  In a solution of 7.8 g benzene (C₆H₆) and 46.0 g toluene (C₆H₅CH₃), the mole-fraction of benzene is

•   1/2
•   1/3
•   1/5
•   1/6

Solution
Mole fraction of C6H6 = (7.8/78) /(7.8/ 78 + 46 /92) = 1/6

Q2. A thermometer which can be used only for accurate measurement of small differences in temperature is known as a:

•   Beckmann thermometer
•   Contact thermometer
•   Clinical thermometer
•   Platinum resistance thermometer

Solution
Beckmann thermometers do not read actual b. p. or f. p., but they give b. p., f. p. values on their scale.

Q3.   A 0.001 molal solution of [Pt(NH₃)₄Cl₄] in water has a freezing point depression of 0.0054 ͦC. If 𝐾_f for water is 1.80, the correct formulation of the above molecule is :

•    [Pt(NH₃)4Cl₃]Cl
•   [Pt(NH)₄Cl₂]Cl₂
•   [Pt(NH₃)₄Cl]Cl₃
•   [Pt(NH₃)₄Cl₄

Solution
[Pt(NH3)4Cl4] =Gives 𝑛 moles of ions on complete ionization, 𝑖.𝑒.,α = 1 ∆𝑇 = 𝐾𝑓 × molality × (1 − 𝛼 + 𝑛𝛼) 0.0054 = 1.80 × 0.001 × (𝑛)

 ∴ 𝑛 = 3 Thus, [Pt(NH₃)₄ Cl₂]Cl₂ ⟶ [Pt(NH₃)₄Cl₂]²+ + 2Cl^- 1 0 0 1 0 2

 ∴ 𝑛 = 3

Q4.  The molal elevation constant for water is 0.52. What will be the boiling point of 2 molar sucrose solution at 1 atm pressure? (Assume b.p. of pure water is 100℃)

•   101.04℃
•   100.26℃
•   100.52℃
•   99.74℃

Solution
∆𝑇_b = 𝑘_b × 𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 for dilute solution molarity=molality=2(given) and 𝑘_b = 0.52(given) 

 ∴ ∆𝑇𝑏 = 0.52 × 2 = 1.04℃ 

Now , ∆𝑇_b = boiling point of solution – boiling point of solvent ( i.e., 𝐻2𝑂) 

 ∴ boiling point of solution =∆𝑇𝑏 + 𝑏.𝑝𝑡 𝑜𝑓 𝐻₂𝑂 =1.04+100

 =101.04℃

Q5. At certain temperature a 5.12% solution of cane sugar is isotonic with a 0.9% solution of an unknown solute. The molar mass of solute is

•   60
•   46.17
•   120
•   90

Solution
  “Solutions having same osmotic pressure are called isotonic solutions.” The osmotic pressure is given as ∴ 𝜋 = 𝑤𝑏𝑅𝑇 𝑉𝑀𝐵 𝜋 (cane sugar)=𝜋 (unknown solute) 5.12/ 342 = 0.9 /𝑀 M= 342×0.9/5.12 =60 .

Q6.  Molarity of 0.2 N H₂SO₄ is

•   0.1
•   0.2
• 0.3
•   0.4

Solution
Normality of acid=molarity ×basicity 0.2 = 𝑀 × 2 ∴ 𝑀 = 0.2/2 = 0.1

Q7. 3.0 molal NaOH solution has a density of 1.110 g/mL. The molarity of the solution is  

 

•   3.9732
•   2.9732
•   1.9732
•   0.9732

Solution

Molarity(𝑚) = 𝑀 /(1000𝑑 − 𝑀𝑀′)× 1000

 Where 𝑀′ =molar mass of solute 

 3 = 𝑀 × 1000/(1000 × 1.11 − 𝑀 × 40) 

 1000𝑀 = 3330 − 120 M 

1120 𝑀 = 3330 

 𝑀 = 3330 /1120

 = 2.9732

Q8. How many gram of NaOH will be required to prepare 500 g solution containing 10%𝑤/𝑤 NaOH solution

•   100 g
• 50g
•   0.5g
•   5.0

Solution
Given, in 100 g of solution NaOH present = 10 g 

 ∴ In 500 g of solution NaOH present = (10 × 500) /100 =50 g 

So, 50 go NaOH will be required to prepare 500 g 10% 𝑤/𝑤 NaOH solution.

Q9. The molal elevation/depression constant depends upon

•   Nature of solvent
•   Nature of solute
•   Temperature
•   ∆𝐻 solution

Solution
It is a characteristic of given solvent

Q10.  The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and nonelectrolyte solid weighing 2.175 g is added to 39.08 g of benzene. If the vapour pressure of the solution is 600 mm of Hg, what is the molecular weight of solid substance?  

 

•   49.50
•   59.60
•   69.60
•   79.82

Solution
Given, vapour pressure of benzene, 

                                             𝑝^o=640 mm Hg 

Vapour pressure of solution, 

                                            p =600 mm

 Hg Weight of solute, w = 2.175 g 

Weight of benzene, W = 39.08 g 

 Molecular weight of benzene, M=78 g 

 Molecular weight of solute, m =? 

 According to Raoult’s law, 

𝑃^o−𝑃 / 𝑃^o = 𝑤×𝑀 / 𝑚×𝑊 

 640−600 / 640 = 2.175×78 / 𝑚×39.08

 40 / 640 = 2.175×78 /𝑚×39.08 

m = 16×2.175×78 / 39.08 

 m =69.60