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 vapour pressure of a solution is proportional to :
  •   Mole fraction of solute
  •   1/(mole fraction of solute
  •   Mole fraction of solvent
  •   None of the above
From Raoult’s law : 𝑃0−𝑃s/𝑃0 = 𝑁1/(𝑁1+𝑁2
1 − 𝑃s /𝑃0 = 𝑁1/ (𝑁1 + 𝑁2
or 𝑃s /𝑃0 = 𝑁2/(𝑁1+𝑁2)
 π‘–.𝑒., 𝑃s = 𝑃0 =𝑁2 /(𝑁1 + 𝑁2)

 π‘ƒs = 𝑃0 × mole fraction of solvent

Q2. At low concentrations, the statements that equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true for
  •   Solutions of non-electrolytes only
  •   Solutions of electrolytes only
  •   All solutions
  •   None of the above
Equal osmatic pressure only applicable to nonelectrolytes solution at low concentration

Q3.   When a solute is added in two immiscible solvents, it distributes itself between two liquids so that its concentration in first liquid is 𝑐1 and that in the second liquid is 𝑐2. If the solute forms a stable trimer in the first liquid, the distribution law suggests that :
  •    3𝑐1 = 𝑐2
  •   𝑐1/3√𝑐2 = constant
  •   𝑐1/3 = 𝑐2
  •   𝑐2/3√𝑐1 = constant
This is the mathematically modified form of distribution law when solute undergoes association in either of the solvent.

Q4.  Solution A contains 7 g/L of 𝑀𝑔𝐢𝑙2 and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of
  •   Solution A is greater than B
  •   Both have same osmotic pressure
  •   Solution B is greater than A
  •   Cannot be determine
Osmotic pressure is a colligative property . More the number of particles (or ions) in solution, more will be osmotic pressure. Nacl solution 
 Given , mass of NaCl =7 g V=1L 
∴ Concentration = mass / mol.mass = 7/58.5 = 0.119 𝑀 
 NaCl dissociates as follows NaCl → π‘π‘Ž+𝐢𝑙−(2 ions) 
∴ Concentration of ions in solution =2× 0.119 𝑀 =0.0238 M MgCl solution
 Given, mass of𝑀𝑔𝐢𝑙2=7g, V=1L 
∴ Concentration== π‘šπ‘Žπ‘ π‘  π‘šπ‘œπ‘™.π‘šπ‘Žπ‘ π‘  = 7 95 = 0.0747
 π‘€π‘”𝐢𝑙2 dissociates as follows 
 MgC𝑙2 → 𝑀𝑔2 + 2𝐢𝑙- (3 ions) 
∴ Concentration of ions in solution =3×0.074 M =0.222 M 
 ∵ Number of particles in solution B (NaCl)are more than in solution A .
 ∴ Osmotic pressure of solution B (NaCl) will be more than solution A .

Q5. The normality of mixture obtained by mixing 100 mL of 0.2 M 𝐻2𝑆𝑂4 + 100 mL of 0.2 M NaOH is
  •   0.2
  •   0.01
  •   0.1
  •   0.3
 H2SO4 - V=100mL, N=0.2 M 
NaOH - V=100mL, N=0.2 M 
Milliequivalent of 𝐻2𝑆𝑂4 = 100 × 0.2 × 2 = 40 (∴ It is dibasic acid) 
 Milliequivalent of NaOH =100 × 0.1 × 2 = 20 
 ∴ Moilliequivalent Of 𝐻2𝑆𝑂4 left =40-20=20 
Total volume = 100mL+100mL=200mL 
Normality of 𝐻2𝑆𝑂4 (left0)= 20/200 =0.1 N .

Q6.  Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1:1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are
  •   250 mm, 550 mm
  •   350 mm, 450 mm
  • 350 mm, 700 mm
  •   550 mm, 250 mm
In Ist case, 
When two liquids X and Y are mixed in the molar ratio 1:1. 
 Moles of X = 1 Moles of Y =1 
 Mole fraction of X (𝜘X)= 1 2 
Mole fraction of Y (𝜘Y)= 1 2 
We know that 
 p=𝑝X 𝜘X+ 𝑝Y 𝜘Y ( p =total pressure of mixture) 
 400 = 1/2 𝑝X + 1/2 𝑝Y  
 400 × 2 = 𝑝X + 𝑝Y …(i) 
 For case IInd
 When liquids are mixed in the molar ratio of 1:2, 
 Mole fraction of X =1 
 Mole fraction of Y = 2 
 Mole fraction of X (𝜘X) = 1/3 
Mole fraction of Y (𝜘Y) = 2/3 
𝑃 = 𝑝X 𝜘X + 𝑝Y 𝜘Y 
350 = 1/3 𝑝X 2/3 π‘π‘Œ
  350 × 3 = 𝑝X + 2𝑝Y …(ii) 
 From Eqs (i) and (ii) , we get 
 π‘X = 550π‘šπ‘š 𝑝Y ° = 250 π‘šπ‘š

Q7. Iodine was added to a system of water and CS2. The concentration of I2 in water and CS2were found to be 𝐢1/𝐢2 respectively. The ratio of 𝐢1/𝐢2 will change if :  

  •   More I2 is added
  •   More CS2 is added
  •   More water is added
  •   Temperature is changed

𝐾 = 𝑐1/𝑐2 is constant for a particular solute in a given solvent-solvent system at constant temperature.

Q8. Which of the following solutions will have the highest boiling point ? NaOH solution
  •   Camphor
  • Naphthalene
  •   Benzene
  •   Water
The molal depression constant (π‘˜f) for camphor is maximum. Hence depression of freezing point (∆𝑇f) will be maximum for camphor =3.3%

Q9. Pressure cooker reduces cooking time for food because

  •   Boiling point of water involved in cooking is increased
  •   Heat is more evenly distributed in the cooking space
  •   The higher pressure inside the cooker crushes the food material
  •   Cooking involves chemical changes helped by a rise in temperature
Due to higher pressure inside the boiling point elevated

Q10.  The molal boiling point constant of water is 0.53 ͦC. When 2 mole of glucose are dissolved in 4000 g of water, the solution will boil at :  


  •   100.53 ͦC
  •   101.06 ͦC
  •   100.265 ͦC
  •   99.47 ͦC
∆𝑇 = (1000 × πΎ × π‘€) /π‘š.π‘Š ; 
 ∴ ∆𝑇 = (1000 × 0.53 × 2 )/4000 = 0.265, 
 ∴ 𝑇b = 100 + 0.265 = 100.265 ͦC

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