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SOLUTIONS QUIZ 19

Dear Readers,


The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesnΓ’€™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.  The difference between the boiling point and freezing point of an aqueous solution containing sucrose (mol wt. = 342 gmol-1) in 100 g of water is 105.0∘𝐢. If 𝐾fand 𝐾b of water are 1.86 and 0.51 K kg mol-1 respectively, the weight of sucrose in the solution is about
  •   1 M solution of glucose
  •   0.05 M solution of glucose
  •   6% solution of glucose
  •   25% solution of glucose
Solution
Molarity of urea = (6/60)/(100 /1000) = 1𝑀 Hence, 
1 M solution of glucose is isotonic with 6% urea solution.

Q2. The mass of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of urea in the same quantity of water is
  •   1 g
  •   3 g
  •   6 g
  •   18 g
Solution
𝑝−𝑝S / 𝑝 = 𝑀1𝑀2 /𝑀2𝑀1 
To produce same lowering of vapour pressure, 𝑝−𝑝s/𝑝 will be same for both cases. 
 So, π‘Š(Glucose)×18/ 50×180 = π‘Š(urea)×18 /50×60 
π‘Š(Glucose) =weight of glucose
 π‘Š(urea) =weight of urea or 
π‘Š (Glucose)×18 50×180 = 1×18 /50×60 
 π‘Š(urea)=3


Q3.   The ratio of vapour pressure over solution phase on mixing two immiscible liquids is equal to :
  •    Ratio of their weights in mixture
  •   Ratio of their mol. weights
  •   Ratio of their moles in liquid phase
  •   Ration of their moles in vapour phase
Solution
𝑃´A = 𝑃A0.𝑋A and 
𝑃´A = 𝑃M ∙ 𝑋´A 
𝑃´B = 𝑃M ∙ 𝑋´B 
 ∴ 𝑃´A /𝑃´B = 𝑋´A /𝑋´B = (𝑛A)V (𝑛B)𝑉

Q4.  Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
  •   1.14 mol π‘˜π‘”-1
  •   3.28 mol π‘˜π‘”-1
  •   2.28 mol π‘˜π‘”-1
  •   0.44 mol π‘˜π‘”-1
Solution
Molality ( m ) = 𝑀 /1000𝑑−𝑀𝑀1× 100 
 M = Molarity 
𝑀1 = Molecular mass 
d = density 
= 2.05 /(1000 × 1.02)−(2.05 × 60) ×100 
 =2.28 mol kg-1



Q5. Two solutions of KNO3 and CH3COOH are prepared separately. Molarity of both is 0.1 M and osmatic pressures are 𝑝1 and 𝑝2respectively. The correct relationship between the osmatic pressures is
  •   𝑝1= 𝑝2
  •   𝑝1 > 𝑝2
  •   𝑝2 > 𝑝1
  •   𝑝1 /𝑝1 + 𝑝2 + 𝑝2/𝑝1 + 𝑝2
Solution
  KNO3 dissociates completely while CH3COOH dissociates to a small extent hence, 𝑝1 > 𝑝2

Q6.  An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N sodium hydroxide required to completely neutralise 10 mL of this solution is
  •   40 mL
  •   20 mL
  • 10 mL
  •   4 mL
Solution
π‘Š = 𝑁𝐸 𝑉 / 1000
 π‘ = π‘Š×1000/ 𝐸×𝑉 
= 6.3×1000 /63×250 
 =0.4N 
 π‘1𝑉1 = 𝑁2𝑉2 
 0.1 × π‘‰1 = 0.4 × 10 
 π‘‰1 = 0.4 × 10 /0.1
 π‘‰1 = 40 mL


Q7. The vapour pressure of water at 20∘C is 17.5 mmHg. If 18 g of glucose (𝐢6𝐻12𝑂6) is added to 178.2 g of water at 20∘𝐢, the vapour pressure of the resulting solution will be  

 
  •   17.675 mmHg
  •   15.750 mmHg
  •   15.750 mmHg
  •   17.325 mmHg
Solution

Moles of glucose = 18/180 = 0.1 
Moles of 𝐻2𝑂 = 178.2 /18 9.9 
According to Raoult’s law 
𝑃°−𝑃s /𝑃° = 𝑋solute 
17.5−𝑃s / 17.5 =0.1/10 so, 
 π‘ƒs = 17.325mm Hg

Q8. An azeotropic mixture of two liquids has boiling point lower than either of them, when it NaOH solution
  •   Shows a negative deviation from Raoult’s law
  • Shows no deviation from Raoult’s law
  •   Shows positive deviation from Raoult’s law
  •   Is saturated
Solution
(i)Azeotropic mixtures having boiling point less than either of the two pure components show positive deviation from Raoult’s law. (ii) Azeotropic mixtures having boiling point more than either of two pure components show negative deviation from Raoult’s law.


Q9. T The unit of molality is

  •   mol 𝐿-1
  •   mol π‘˜π‘”-1
  •   π‘šπ‘œπ‘™-1 𝐿-1
  •  mol L
Solution
Unit of molality mole per kilogram (mol kg-1).

Q10.  The lubricating action of an oil is more if it possess :  

 

  •   High vapour pressure
  •   Low vapour pressure
  •   High surface tension
  •   High density
Solution
Substances of high vapour pressure (𝑒.g., gasoline) evaporates more quickly than substances of low vapour pressure (𝑒.g., motor oil).

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SOLUTIONS QUIZ 19
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