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SOLUTIONS Quiz-14

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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.  A solute when distributed between two immiscible phases remains associated in phase II and dissociated in phase I. If Ξ± is the degree of dissociation and 𝑛 is the number of molecules associated then :
  •   𝐾 = 𝑐I/𝑐II
  •   𝐾 = 𝑐I/√𝑐II(1 − Ξ±)
  •   𝐾 = 𝑐I / 𝑐II (1 − Ξ±)
  •   𝐾 = 𝑐I(1 − Ξ±)/ n√𝑐II
Solution
This is the mathematically modified form of distribution law when solute undergoes association in either of the solvent

Q2. Saturated solution of NaCl on heating becomes :
  •   Super saturated
  •   Unsaturated
  •   Remains saturated
  •   None of these
Solution
On heating solubility of NaCl increases.


Q3.   At 25∘C, the total pressure of an ideal solution obtained by mixing 3 moles of ‘ A ’ and 2 moles of ‘ B ’, is 184 torr. What is the vapour pressure (in torr) of pure ‘ B ’ at the same temperature? (Vapour pressure of pure ‘ A ’ at 25∘C is 200 torr)
  •   180
  •   160
  •   16
  •   100
Solution
P = 𝑝A ∘ ( 𝑛A /𝑛A + 𝑛B ) + 𝑝B ∘ ( 𝑛B 𝑛A + 𝑛B ) 184 = 200( 3/(3 + 2) )𝑝B ∘ ( 2/(3 + 2 ) 184 = 200 × 3/5 + 𝑝B∘ × 2/5 184=200+𝑝B ∘ 2/5 64=𝑝B ∘ 2/5 𝑝B ∘ =64×5 2 =160 torr

Q4.  At 40 ͦC the vapour pressures of pure liquids, benzene and toluene, are 75 torr and 22 torr respectively. At the same temperature, the partial vapour pressure of benzene in a mixture of 78 g benzene and 46 g toluene in torr assuming the ideal solution should be :
  •   50
  •   25
  •   375
  •   53.5
Solution
𝑃M= 𝑃´Benzene + 𝑃´Toluene
  𝑃M = 75 × (78/78) / (78/78 + 46/92) + 22 × (46/82) / (78 /78 + 46/92) 
 ?PM = 75 × 2 3 + 22 × 1 2 × 2 3 = 50 + 7.3 = 57.3
 Also 𝑃´?A = 50


Q5. Solutions A, B , C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl2 and 0.1 M AlCl3. Which one of the following pairs is isotonic?
  •   A and B
  •   B and C
  •   A and D
  •   A and C
Solution
  Isotonic solutions have same molar concentration of solute particles in solution.
 Molar concentration of particles in solution are 0.1 M in glucose, 2×0.05 M in NaCl, 3×0.05 in BasCl2 and 4×0.1 in AlCl3. Therefore, 0.1 M glucose and 0.05 in M NaCl solutions are isotonic .

Q6.  𝑝A and 𝑝B are the vapour pressure of pure liquid components 𝐴 and 𝐡 respectively of an ideal binary solution. If π‘₯𝐴 represents the mole fraction of component 𝐴, the total pressure of the solution will be :
  •   𝑝B + π‘₯A(𝑝B − 𝑝A)
  •   𝑝B + π‘₯A(𝑝A − 𝑝A)
  • 𝑝A + π‘₯A(𝑝A − 𝑝A)
  •   𝑝A + π‘₯A(𝑝A − 𝑝B)
Solution
𝑝M = 𝑝´A + 𝑝´B 
 = 𝑝A ∙ π‘₯B + 𝑝B ∙ π‘₯B         (∵ 𝑝´a = 𝑝A ∙ π‘₯A
 = 𝑝A ∙ π‘₯A + 𝑝A(1 − π‘₯A)   (∵ π‘₯A + π‘₯B = 1) 
 = 𝑝b + π‘₯a(𝑝A − 𝑝B)


Q7. The vapour pressure of water at 23℃ is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution?  

 
  •   19.0
  •   19.602
  •   19.402
  •   19.202
Solution

Given 𝑝𝑠 =19.8 mm 𝑛A = 0.1 𝑛B = 178.2 18 = 9.9 According to Raoult’s law 𝑝s−𝑝 / 𝑝s = 𝑛𝐴 𝑛𝐴+𝑛𝐡 19.8−𝑝 / 19.8 = 0.1 /(9.9+0.1) or 198-10 p = 19.8 ×0.1 10 p = 198-1.98 10 p =196.02 p = 19.602 mm


Q8. A solution of sucrose (Molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of solution, what is its osmotic pressure (R=0.082 L atmK-1 mol-1) at 273 K? NaOH solution
  •   0.01 M Na2SO4
  • 0.01 M 𝐾𝑁𝑂3
  •   0.015 M urea
  •   0.015 M glucose
Solution
Boiling point =𝑇0(π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘) + ∆𝑇b(Elevation in b.p.)
 ∆𝑇b = π‘šπ‘–π‘˜b 
 where, m is the molality i.e., the van’t Hoff factor ( i ) 
=[1+(y-1) x ] 
 π‘˜π‘ = π‘šπ‘œπ‘™π‘Žπ‘™ elevation constant. 
 Thus, ∆𝑇b ∝ π‘–π‘š
 Assume 100% ionisation 
 (a) π‘šπ‘–(π‘π‘Ž2𝑆𝑂4) = 0.01 × 3 = 0.03 
 (b) π‘šπ‘–(𝐾𝑁𝑂3) = 0.01 × 2 = 0.02
 (c) mi (urea)=0.015
 (d) mi (glucose)=0.015


Q9. Lowering in vapour pressure is the highest for:

  •   0.2 π‘š urea
  •   0.1 π‘š glucose
  •   0.1 π‘š MgSO4
  •   0.1 π‘š BaCl2
Solution
𝑃0−𝑃s / 𝑃0 = molality × (1 ⎯ 𝛂 + π‘₯𝛂 + y𝛂) 
The value of 𝑃0 − 𝑃s is maximum for BaCl2.

Q10.  The 𝐾 for I2 between CS2 and H2O is 588 in favour of CS2. One litre of aqueous solution containing 1 g of I is shaken with 50 mL of CS2. What will be the amount of I2 in aqueous layer?  

 

  •   0.035 g
  •   0.010 g
  •   0.05 g
  •   0.04 g
Solution
𝐾 = 588 = π‘₯/50 /(1−π‘₯) /1000
 ∴ π‘₯ = 0.965 g Where, π‘₯ is amount of I2 in CS2.
 Thus, it aqueous layer I2 = 1 − 0.965 = 0.035 g

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SOLUTION QUIZ 14
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