SOLUTIONS QUIZ 20

 

SOLUTIONS QUIZ 20

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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

Solution
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

Solution
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

Solution
𝐾 = 420 = 5 − 𝑥 /𝑥 ∴ 𝑥 = 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

Solution
𝐻₃𝑃𝑂₃ is a dibasic acid (containing two ionisable protons attached to O directly). 

 𝐻₃𝑃𝑂₃ ⇌ 2𝐻+ + 𝐻𝑃𝑂₃^2- 

∴ 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

Solution
  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

Solution
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℃

Solution

△ 𝑇_f = 𝑖𝑘_f𝑚 𝑤ℎ𝑒𝑟𝑒

 △ 𝑇^f = depression in freezing point i=van,

t Hoff factor m = molality and 

and 𝑘_f= freezing point depression constant For 0.01 molal NaCl solution 

 0.37 = 2 × 𝑘_f × 0.01 

∴ 𝑘_f = 0.37 2×0.01 -----(i) 

 For 0.02 molal urea solution 

 △ 𝑇_f = 1 × 𝑘_f × 0.02 

∴ 𝑘_f = △𝑇_f 0.02 -----(ii) 

From Eqs (i) and (ii) 

0.37 2×0.01 = △𝑇_f 0.02 

 △ 𝑇_f = 0.37×0.02 2×0.01 

∴ △ 𝑇_f = 0.37∘𝐶

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

Solution
𝑀 = 𝑊/ 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

Solution
𝑝⁰ − 𝑝_s / 𝑝₀ = 𝑤 ×𝑀 𝑚 × 𝑊 

 143 − 𝑝_s / 143 = 0.5 /65 × 154/ 1.58 × 100 

 [∵ molecular weight of CCl₄ = 154 and weight=density×volume] 

143 − 𝑝_s = 1.07 ⇒ 𝑝_s = 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

Solution
𝑝⁰ − 𝑝_s /𝑝⁰ = 𝑤 /𝑚 × 𝑀/ 𝑤 

 0.30 mm /17.54 mm = 20/ 𝑚 × 18 /100 

⇒ 𝑚 = 20 × 18 × 17.54/ 0.30 × 100 = 210.48