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 weight of H2C2O4 ∙ 2H2O required to pressure 500 mL of 0.2 N solution is
  •   63 g
  •   6.3 g
  •   0.63 g
  •   126 g
Eq. wt. of H2C2O4 ∙ 2H2
 = 2 + 24 + 64 + 2(2 + 16) /2 = 63  
 π‘€ = 𝑁𝐸𝑉 / 1000 = (0.2 × 63 × 500)/ 1000 = 6.3 g

Q2. The elevation of boiling point method is used for the determination of molecular weight of:
  •   Non-volatile and soluble solute
  •   Non-volatile and insoluble solute
  •   Volatile and soluble solute
  •   Volatile and insoluble solute
To show colligative properties solute should be non-volatile and soluble in given solvent.

Q3.   50 π‘π‘š3of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding 50 π‘π‘š3of NaOH. The remaining titration is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is
  •    12 π‘π‘š3
  •   10 π‘π‘š3
  •   25 π‘π‘š3
  •   10.5 π‘π‘š3
When 0.1 n NaOH is used, 
       𝑁1𝑉1 = 𝑁1𝑉2 
 (For HCl) (For KOH) 
0.2𝑁 × π‘‰1 = 50 × 0.1𝑁 
 π‘‰1 = 50 ×0.1 /0.2 = 25π‘π‘š3 
 When 0.5 N KOH is used, 
                       𝑁1𝑉1 = 𝑁3𝑉3 
 (For remaining HCl) (for KOH) 
 0.2𝑁 × 25 = 0.5 𝑁 × π‘‰3 𝑉3 = (0.2 ×25) / 0.5 = 10π‘π‘š3

Q4.  Dissolution of a solute is an exothermic process if :
  •   Hydration energy > lattice energy
  •   Hydration energy < lattice energy
  •   Hydration energy = lattice energy
  •   None of the above
∆𝐻solution = ∆𝐻hydration + ∆𝐻lattice energy
 ∆𝐻h = −ve 
∆𝐻l = +ve

Q5. Which of the following shows maximum depression in freezing point?
  •   𝐾2𝑆𝑂4
  •   NaCl
  •   Urea
  •   glucose
  Depression in freezing point is a colligative property. It depends on number of particles. More the number of particles, more will be depression in freezing point. 
 1. 𝐾2𝑆𝑂4 → 2𝐾+ + 𝑆𝑂4 2− It gives 3 particles. 
2. π‘π‘ŽπΆπ‘™ → π‘π‘Ž+ + 𝐢𝑙− It gives 2 particles. 
 3. Urea→ No dissociation 4. Glucose → No dissociation. 
∴ 𝐾2𝑆𝑂4 produces maximum number of particles 
∴ 𝐾2𝑆𝑂4 has maximum depression in freezing point. .

Q6.  Water will boil at 101.5 ͦC at which of the following pressure?
  •   76 cm of Hg
  •   76 mm of Hg
  • > 76 cm of Hg
  •   < 76 cm of Hg
Water boils at higher temperature than its b. p. if atmosphere pressure is more than 1 atm.

Q7. The vapour pressure of pure liquid A is 0.80 atm. When a non-volatile B is added to A its vapour pressure drops to 0.60 atm. The mole fraction of B in the solution is  

  •   0.125
  •   0.25
  •   0.5
  •   0.75

According to the Raoult’s law the relative lowering vapour pressure which is produced by dissolving a non-volatile solute in a solvent is equal to mole fraction of the solute. 
 π‘ƒ−𝑃s/𝑃 = 𝑋𝐡 
 where, p= vapour pressure of solvent 
 π‘ƒ/𝑠 = vapour pressure of solution 
𝑋𝐡 = mole fraction of 𝐡 
Given, P=0.80 atm 𝑃s = 0.60 atm 
 ∴ 𝑋B = 0.80−0.60/0.80
 = 0.20/0.80 
 = 0.25

Q8. One gram of silver gets distributed between 10 cm3 of molten zinc and 100cm3of molten lead at 8000℃. The percentage of silver still left in the lead layer in approximately NaOH solution
  •   2
  • 5
  •   3
  •   1
𝐾𝐷 = π‘π‘œπ‘›π‘π‘’π‘›π‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ 𝑋 𝑖𝑛 π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘ / 𝐴 π‘π‘œπ‘›π‘π‘’π‘›π‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ 𝑋 𝑖𝑛 π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘ 𝐡 
 Concentration of Ag in 10 π‘π‘š3 Zn=x 
Concentration of Ag in 100 π‘π‘š3 Pb=1−π‘₯ /10 
Concentration of Ag in 10 π‘π‘š3 Pb=1−π‘₯ / 10
 300= π‘₯×10 /(1−π‘₯) π‘œπ‘Ÿ
 π‘₯ = 300/ 100 = 0.967 
 = 97% 
Concentration of Ag in zinc =1 -0.967=0.033 =3.3%

Q9. How many moles of 𝐴𝑙2 (𝑆𝑂4 )3 would be in 50 g of the substance?

  •   0.083 mol
  •   0.952 mol
  •   0.481 mol
  •   0.140 mol
Moles = mass molecular weight 
 Given, mass of 𝐴𝑙2 (𝑆𝑂4)3 = 50 g 
molecular mass of 𝐴𝑙3 (𝑆𝑂4)3 = 342 
∴ Moles of 𝐴𝑙2 (𝑆𝑂4)3 = 50 /342 = 0.14 mol

Q10.  The distribution law was given by :  


  •   Henry
  •   Nernst
  •   van’t Hoff
  •   Ostwald
It is therefore also known as Nernst distribution law

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