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.  In which ratio of volume 0.4 M HCl and 0.9 M HCl are to be mixed such that the concentration of the resultant solution becomes 0.7 M ?
  •   Air
  •   Brass
  •   Amalgam
  •   Benzene in water
Benzene in water

Q2. Boiling point of water is defined as the temperature at which :
  •   Vapour pressure of water is equal to one atmospheric pressure
  •   Bubbles are formed
  •  Steam comes out
  •   None of the above
It is the definition of boiling point.

Q3.   A solution of 5 g of iodine in CS2 was shaken with the same volume of water. The amount of iodine in water is : (Given 𝐾 in favour of CS2 = 420
  •    0.119 g
  •   0.0119 g
  •   0.00119 g
  •   1.19 g
𝐾 = 420 = 5 − π‘₯ /π‘₯ ∴ π‘₯ = 0.0119 g

Q4.  To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (𝐻3𝑃𝑂3),the volume of 0.1 M aqueous KOH solution required is
  •   10 mL
  •   20 mL
  •   40 mL
  •   60 mL
𝐻3𝑃𝑂3 is a dibasic acid (containing two ionisable protons attached to O directly). 
 π»3𝑃𝑂3 ⇌ 2𝐻+ + 𝐻𝑃𝑂32- 
∴ 0.1 𝑀 𝐻3𝑃𝑂3 = 0.2 𝑁𝐻3 𝑃𝑂3 
 and 0.1 M KOH = 0.1N KOH 
   𝑁1𝑉1 = 𝑁2𝑉2 
 (KOH) (𝐻3𝑃𝑂3
0.1𝑉1 = 0.2 × 20 
𝑉1 = 40mL

Q5. At high altitude the boiling of water occurs at low temp. because :
  •   Atmospheric pressure is low
  •   Temperature is low
  •   Atmospheric pressure is high
  •   None of the above + 𝑝2/𝑝1 + 𝑝2
  The boiling occurs at lower temperature if atmospheric pressure is lower than 76 cm Hg.

Q6.  The movement of solvent molecules through a semipermeable membrane is called
  •   Electrolysis
  •   Electrophoresis
  • Osmosis
  •   Cataphoresis
The phenomenon in which, when two solutions of different concentration (one may be solvent) are kept separated by semipermeable membrane, the solvent molecules start flowing from dilute solution to concentrate solution. This is called osmosis. Osmosis is a slow process and keeps on happening until the concentration of both solutions become equal.

Q7. The freezing point of water is depressed by 0.37℃ in a 0.01 mol NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by  

  •   0.37℃
  •   0.74℃
  •   0.185℃
  •   0℃

△ 𝑇f = π‘–π‘˜fπ‘š π‘€β„Žπ‘’π‘Ÿπ‘’
 △ 𝑇f = depression in freezing point i=van,
t Hoff factor m = molality and 
and π‘˜f= freezing point depression constant For 0.01 molal NaCl solution 
 0.37 = 2 × π‘˜f × 0.01 
∴ π‘˜f = 0.37 2×0.01 -----(i) 
 For 0.02 molal urea solution 
 △ 𝑇f = 1 × π‘˜f × 0.02 
∴ π‘˜f = △𝑇f 0.02 -----(ii) 
From Eqs (i) and (ii) 
0.37 2×0.01 = △𝑇f 0.02 
 △ 𝑇f = 0.37×0.02 2×0.01 
∴ △ 𝑇f = 0.37∘𝐢

Q8. If 5.85 g NaCl (molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 L, the molarity of the solution will be NaOH solution
  •   0.1
  • 0.2
  •   0.3
  •   0.4
𝑀 = π‘Š/ mol.wt.×𝑉(L) = 5.85 /58.5 × 0.5 = 0.2 M 
(a)6g of NaOH/100 mL 
(b)0.5 M H2SO4 
𝑁 = 𝑀 × basicity = 0.5 × 2 = 1.0 
(c)𝑁 phosphoric acid Normality=1 
(d)8 g of KOH/L Normality = (strength in g/L) /equivalent weight = 8 /56 = 0.14 N

Q9. Vapour pressure of CCl4 at 25℃ is 143 mm of Hg and 0.5 g of a non-volatile solute (mol. wt=65) is dissolved in100 mL CCl4. Find the vapour pressure of the solution. (Density of CCl4 = 1.58 g/cm2)

  •   94.39 mm
  •   141.93 mm
  •   134.44 mm
  •  199.34 mm
𝑝0 − 𝑝s / 𝑝0 = 𝑀 ×𝑀 π‘š × π‘Š 
 143 − 𝑝s / 143 = 0.5 /65 × 154/ 1.58 × 100 
 [∵ molecular weight of CCl4 = 154 and weight=density×volume] 
143 − 𝑝s = 1.07 ⇒ 𝑝s = 141.93 mm

Q10.  The vapour pressure of water at 20℃ is 17.54 mm. When 20 g of a non-ionic, substance is dissolved in 100 g of water, the vapour pressure is lowered by 0.30 mm. What is the molecular mass of the substance?  


  •   200.8
  •   206.88
  •   210.5
  •   215.2
𝑝0 − 𝑝s /𝑝0 = 𝑀 /π‘š × π‘€/ 𝑀 
 0.30 mm /17.54 mm = 20/ π‘š × 18 /100 
⇒ π‘š = 20 × 18 × 17.54/ 0.30 × 100 = 210.48

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