## SOLUTIONS Quiz-15

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’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 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−𝑃s /𝑃 = 𝑋2
𝑃−𝑃s/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 (𝑘f = 1.86)
•   135.0
•   172.0
•   90.0
•   180.0
Solution
M= 1000×𝑘f×𝑤 / ∆𝑇f×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 𝐾f and 𝐾b 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
∆𝑇b / ∆𝑇f = 𝐾b / 𝐾f
∴ ∆𝑇f = 𝐾f /𝐾b × ∆𝑇b = 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
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃A ∘𝑋A + 𝑃B ∘𝑋B
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,
𝑃𝑡𝑜𝑡𝑎𝑙 = 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
𝑃M = 𝑃0𝐶5𝐻12 .𝑋0𝐶5𝐻12 + 𝑃0𝐶6𝐻14 .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 ## Want to know more

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SOLUTIONS QUIZ 15