SOLUTIONS QUIZ 12

 

  SOLUTIONS Quiz-12

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advanced. .

Q1.  The amount of urea dissolved in 500 cc of water (𝐾^f = 1.86℃) to produce a depression of 0.186℃ in the freezing point is

•   9 g
•   6 g
•   3 g
•   0.3 g

Solution
We know that,

 w (mass of solute) = 𝑚×∆T_f×𝑊 / 1000×𝑘_f 

m= mol. wt. of urea (60) 

 ∆𝑇_f = 0.186℃ 

 𝑘_f =1.86°, W =500 g

Q2. The osmotic pressure (At27℃) of an aqueous solution (200 mL) containing 6 g of a protein is 2 × 10^-3atm . If R=0.080 L atm mol^-1 K^-1, the molecular weight of protein is

•   7.2 × 10⁵
•   3.6 × 10⁵
•   1.8 × 10⁵
•   1.0 × 10⁵

Solution
𝜋𝑉 = 𝑤/𝑚𝑅𝑇 M= 𝑤𝑅𝑇 / 𝜋𝑉 Here, w=6 g, 𝜋 = 2 × 10^-3𝑎𝑡𝑚, T=300 K, R=0.080 L-atm mo𝑙^-1 𝐾^-1, V =200 mL =0.2 L M= 6 ×0.080×300 2×10³×0.2 = 3.6 × 10⁵

Q3.  Assuming that sea water is a 3.50 weight per cent aqueous solution of NaCl. What is the molality of sea water?

•   0.062 m
•   0.0062 m
•   0.62 m
•   6.2 m

Solution
3.50 wt% of aqueous solution of NaCl means 100 g of sea water contains 3.50 g NaCl. 

 Water in sea water = 100-3.5 = 96.5 g 

 =0.0965 kg 

 Molality = 3.5 58.5 ×0.0965 = 0.62 m

Q4.  When 25 g of a non-volatile solute is dissolved in 100 g of water, the vapour pressure is lowered by 2.25 × 10^-1mm. If the vapour pressure of water at 20℃ is 17.5 mm, what is the molecular weight of the solute?

•   206
•   302
•   350
•   276

Solution
Given, Weight of non-volatile solute, 

 w= 25 g 

 Weight of solvent, W=100 g 

 Lowering of vapour pressure, 𝑝° − 𝑝𝑠 = 0.225 mm 

 Vapour pressure of pure solvent, 𝑝° = 17.5 mm

 Molecular weight of solvent (𝐻2𝑂), M =18 g 

 Molecular weight of solute, m =? 

 According to Raoult’s law

 𝑝°−𝑝_s 𝑝° = 𝑤×𝑀 / 𝑚×𝑊 

 0.225/17.5 = 25 × 18 / 𝑚 × 100 

 𝑚 =25 × 18 × 17.5 / 22.5 = 350 𝑔

Q5. Van’t hoff factor of 𝐶𝑎(𝑁𝑂₃)₂ is

•   Benzoic acid is an organic solute
•   Benzoic acid has higher molar mass than benzene
•   Benzoic acid gets associated in benzene
•   Benzoic acid gets dissociated in benzene

Solution
  Benzoic acid in benzene exists as a dimer. So, number of molecules decreases and hence, osmotic pressure decreases .

Q6.  Dilute 1 L one molar H₂SO₄ solution by 5 L water, the normality of that solution is

•   0.33 N
•   33.0 N
• 0.11 N
•   11.0 N

Solution
0.33 N

Q7. Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 mL solution is  

 

•   1.50 M
•   2.50 M
•   3.50 M
•   5.01 M

Solution
𝑀 = 𝑤 × 1000 / 𝑚 × 𝑉(mL) = 75.5 × 1000 / 56 × 540 = 2.50 M

Q8. What is the total number of moles of 𝐻₂𝑆𝑂₄ needed to prepare 5.0 L of a 2.0 M solution of 𝐻₂𝑆𝑂₄ ?

•   2.5
• 5.0
•   10
•   20

Number of moles = Molarity × Volume (in L) ⟹ Number of moles of 𝐻₂ 𝑆𝑂⁴ = 2.0 𝑀 × 5.0 𝐿 = 10 moles

Q9. The relative lowering of vapour pressure of a dilute aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is about

•   0.70
•   0.50
•   0.90
•   0.80

Solution
Relative lowering of vapour pressure = mole fraction of solute (Raoult’s law ) 𝑝−𝑝_s 𝑝 = 𝑥₂

Q10.  The osmotic pressure of a solution can be accurately measured in the shortest possible time by :  

 

•   Berkeley and Hartley method
•   Morse and Frazer method
•   Pfeffer method
•   None of the above

Solution
It is more precise and takes minimum time.