SOLUTIONS QUIZ 15

 

SOLUTIONS Quiz-15

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Q1. 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 

 𝑃−𝑃_s/𝑃 = 𝑤𝑀/𝑚𝑊 

 where, 

 w =wt. of solute

M =mol. wt. of solvent 

 m = mol. wt. of solute 

W = wt. of solvent 

 0.0125= 𝑤𝑀 / 𝑚𝑊 

or 

 𝑀 /𝑚𝑊 = 0.0125/18 = 0.00070 

Hence, molality = 𝑤/𝑚𝑊 × 1000 = 0.0007 × 1000 = 0.70

Q2. A solution of 4.5 g of a pure non-electrolyte in 100 g of water was found to freeze at 0.465℃ . The molecular weight of the solute closest to (𝑘_f = 1.86)

•   135.0
•   172.0
•   90.0
•   180.0

Solution
M= 1000×𝑘_f×𝑤 / ∆𝑇_f×𝑊 = 1000×1.86×4.5 / 0.465×100 = 180 𝑔

Q3.   A solution of urea (mol. mass 56) boils at 100.18 ͦC at atmospheric pressure. If 𝐾_f and 𝐾_b for water are 1.86 and 0.512 K molality^-1 respectively, the above solution will freeze at :

•    − 6.54 ͦC
•  6.54 ͦC
•   − 0.654 ͦC
•   0.654 ͦC

Solution
∆𝑇_b / ∆𝑇_f = 𝐾_b / 𝐾_f 

∴ ∆𝑇_f = 𝐾_f /𝐾_b × ∆𝑇_b = 1.86 /0.512 × 0.18 = 0.654 

∴ f.pt.= 0 − 0.654 = −0.654 ͦC

Q4.  Vapour pressure of pure A = 100 torr, moles = 2; vapour pressure of pure B =80 torr, moles = 3. Total vapour pressure of the mixture is

•   440 torr
•   460 torr
•   180 torr
•   88 torr

Solution
𝑃_{𝑡𝑜𝑡𝑎𝑙} = 𝑃_A ∘𝑋_A + 𝑃_B ∘𝑋_B

  where , P = vapour pressure X = mole fraction 

 Total moles of A and B = 5 

 Mole fraction of compound A = 2/5 

Mole fraction of compound B = 3/5 then, 

 𝑃_{𝑡𝑜𝑡𝑎𝑙} = 100 × 2/5 + 80 × 3/5 = 88 torr

Q5. The boiling point of C₆H₆, CH₃OH, C₆H₅NH₂ and C₆H₅NO₂ are 80 ͦC, 65 ͦC, 184 ͦC and 212 ͦC respectively. Which will show highest vapour pressure at room temperature?

•   C₆H₆
•   CH₃OH
•   C₆H₅NH₂
•   C₆H₅NO₂

Solution
  Lower is the b. p. of solvent more is its vapour pressure .

Q6.  In a pair of immiscible liquids, a common solute dissolves in both and the equilibrium is reached. The concentration of solute in upper layer is :

•   Same as in lower layer
•   Lower than the lower layer
• Higher than the lower layer
•   In fixed ratio with that in the lower layer

Solution
𝐾 = 𝑐₁/𝑐₂

Q7. Blood cells retain their normal shapes in solutions which are :  

 

•   Isotonic to blood
•   Hypotonic to blood
•   Hypertonic to blood
•  Equinormal to blood

Solution

Osmosis does not take place if two solutions are isotonic.

Q8. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20 ͦC are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in vapour phase would be : NaOH solution

•   0.786
• 0.549
•   0.478
•   0.200

Solution
𝑃_M = 𝑃⁰_𝐶5𝐻12 .𝑋⁰_𝐶5𝐻12 + 𝑃⁰_{𝐶6𝐻14 .𝑋𝐶6𝐻14 ;} 

_{Thus, 𝑃M = 440 × 1/5 + 120 × 4/5 = 184} 

_{Now, 𝑃C5H12 = 𝑃⁰C5H12 ∙ 𝑋C5H12(𝑙) = 𝑃M ∙ 𝑋C5H12(g)} 

_{∴ 440 × 1 5 = 184 × 𝑋C5H12(g)}  

_{∴ 𝑋C5H12(g) = 0.478}

Q9. The empirical formula of a non-electrolyte is 𝐶𝐻₂𝑂 . A solution containing 6g of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The molecular formula of the compound is

•   𝐶₂𝐻₄𝑂₂
•   𝐶₃𝐻₆𝑂₃
•   𝐶₅H₁₀𝑂₅
•   𝐶₄𝐻₅𝑂₄

Solution
Solutions having same osmotic pressure, at a given temperature, have same concentration. Concentration of compound = concentration of glucose 

6/𝑀×1 = 0.05

 M= 6/0.05 = 120 

Empirical formula mass (𝐶𝐻₂𝑂)= 12+2+16 =30 

𝑛 = molecular mass / empirical formula mass = 120 /30 = 4 

 Hence, molecular formula = (𝐶𝐻₂𝑂)₄ = 𝐶₄𝐻₈𝑂₄

Q10.  A 5.2 molal aqueous solution of methyl alcohol, 𝐶𝐻₃𝑂𝐻, is supplied. What is the mole fraction of methyl alcohol in the solution?  

 

•   1.100
•   0.190
•   0.086
•   0.050

Solution
Molality 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 / 𝑘𝑔 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 5.2 𝑚𝑜𝑙 𝐶𝐻₃𝑂𝐻 𝑘𝑔 (=100𝑔)𝐻₂𝑂 

𝑛₁(𝐶𝐻₃𝑂𝐻) = 5.2 𝑛₂(𝐻₂𝑂) = 1000/18 = 55.56

 ∴ 𝑛₁+𝑛₂ = 5.20 + 55.56 = 60.76 mol

 ∴ 𝑋𝐶𝐻₃𝑂𝐻 = 𝑛₁ /(𝑛₁+𝑛₂) = 5.2 /60.76 = 0.086