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****SOLUTIONS Quiz-14**

**SOLUTIONS Quiz-14**

**Dear Readers,**

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.
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**Q1.**A solute when distributed between two immiscible phases remains associated in phase II and dissociated in phase I. If α is the degree of dissociation and 𝑛 is the number of molecules associated then :

Solution

This is the mathematically modified form of distribution law when solute undergoes association in either of the solvent

This is the mathematically modified form of distribution law when solute undergoes association in either of the solvent

**Q2.**Saturated solution of NaCl on heating becomes :

Solution

On heating solubility of NaCl increases.

On heating solubility of NaCl increases.

**Q3.**At 25∘C, the total pressure of an ideal solution obtained by mixing 3 moles of ‘ A ’ and 2 moles of ‘ B ’, is 184 torr. What is the vapour pressure (in torr) of pure ‘ B ’ at the same temperature? (Vapour pressure of pure ‘ A ’ at 25∘C is 200 torr)

Solution

P = 𝑝

P = 𝑝

_{A}∘ ( 𝑛_{A}/𝑛_{A}+ 𝑛_{B}) + 𝑝_{B}∘ ( 𝑛_{B}𝑛_{A}+ 𝑛_{B}) 184 = 200( 3/(3 + 2) )𝑝_{B}∘ ( 2/(3 + 2 ) 184 = 200 × 3/5 + 𝑝_{B}∘ × 2/5 184=200+𝑝_{B}∘ 2/5 64=𝑝_{B}∘ 2/5 𝑝_{B}∘ =64×5 2 =160 torr

**Q4.**At 40 ͦC the vapour pressures of pure liquids, benzene and toluene, are 75 torr and 22 torr respectively. At the same temperature, the partial vapour pressure of benzene in a mixture of 78 g benzene and 46 g toluene in torr assuming the ideal solution should be :

Solution

𝑃

𝑃

_{M}= 𝑃´_{Benzene}+ 𝑃_{´Toluene}_{ }𝑃

_{M}= 75 × (78/78) / (78/78 + 46/92) + 22 × (46/82) / (78 /78 + 46/92)

?P

_{M}= 75 × 2 3 + 22 × 1 2 × 2 3 = 50 + 7.3 = 57.3 Also 𝑃´?

_{A}= 50**Q5.**Solutions A, B , C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl

_{2}and 0.1 M AlCl

_{3}. Which one of the following pairs is isotonic?

Solution

**Isotonic solutions have same molar concentration of solute particles in solution.****Molar concentration of particles in solution are 0.1 M in glucose, 2×0.05 M in NaCl, 3×0.05 in BasCl**.

_{2}and 4×0.1 in AlCl_{3}. Therefore, 0.1 M glucose and 0.05 in M NaCl solutions are isotonic**Q6.**𝑝

_{A}and 𝑝

_{B}are the vapour pressure of pure liquid components 𝐴 and 𝐵 respectively of an ideal binary solution. If 𝑥𝐴 represents the mole fraction of component 𝐴, the total pressure of the solution will be :

Solution

𝑝

𝑝

_{M}= 𝑝´_{A}+ 𝑝´_{B} = 𝑝

_{A}∙ 𝑥_{B}+ 𝑝_{B}∙ 𝑥_{B}(∵ 𝑝´_{a}= 𝑝_{A}∙ 𝑥_{A}) = 𝑝

_{A}∙ 𝑥_{A}+ 𝑝_{A}(1 − 𝑥_{A}) (∵ 𝑥_{A}+ 𝑥_{B}= 1) = 𝑝

_{b}+ 𝑥_{a}(𝑝_{A}− 𝑝_{B})**Q7.**The vapour pressure of water at 23℃ is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution?

Solution

Given 𝑝𝑠 =19.8 mm 𝑛

Given 𝑝𝑠 =19.8 mm 𝑛

_{A}= 0.1 𝑛_{B}= 178.2 18 = 9.9 According to Raoult’s law 𝑝_{s}−𝑝 / 𝑝_{s}= 𝑛𝐴 𝑛𝐴+𝑛𝐵 19.8−𝑝 / 19.8 = 0.1 /(9.9+0.1) or 198-10 p = 19.8 ×0.1 10 p = 198-1.98 10 p =196.02 p = 19.602 mm**Q8.**A solution of sucrose (Molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of solution, what is its osmotic pressure (R=0.082 L atmK

_{-1}mol

_{-1}) at 273 K? NaOH solution

Solution

Boiling point =𝑇

Boiling point =𝑇

_{0}(𝑠𝑜𝑙𝑣𝑒𝑛𝑡) + ∆𝑇_{b}(Elevation in b.p.) ∆𝑇

_{b}= 𝑚𝑖𝑘_{b} where, m is the molality
i.e., the van’t Hoff factor ( i )

=[1+(y-1) x ]

𝑘𝑏 = 𝑚𝑜𝑙𝑎𝑙 elevation constant.

Thus, ∆𝑇

_{b}∝ 𝑖𝑚 Assume 100% ionisation

(a) 𝑚𝑖(𝑁𝑎

_{2}𝑆𝑂_{4}) = 0.01 × 3 = 0.03 (b) 𝑚𝑖(𝐾𝑁𝑂

_{3}) = 0.01 × 2 = 0.02 (c) mi (urea)=0.015

(d) mi (glucose)=0.015

**Q9.**Lowering in vapour pressure is the highest for:

Solution

𝑃

𝑃

_{0}−𝑃_{s}/ 𝑃0 = molality × (1 ⎯ 𝛂 + 𝑥𝛂 + y𝛂)The value of 𝑃

_{0}− 𝑃_{s}is maximum for BaCl_{2}.**Q10.**The 𝐾 for I

_{2}between CS

_{2}and H

_{2}O is 588 in favour of CS

_{2}. One litre of aqueous solution containing 1 g of I

_{}is shaken with 50 mL of CS

_{2}. What will be the amount of I

_{2}in aqueous layer?

Solution

𝐾 = 588 = 𝑥/50 /(1−𝑥) /1000

𝐾 = 588 = 𝑥/50 /(1−𝑥) /1000

∴ 𝑥 = 0.965 g Where, 𝑥 is amount of I

_{2}in CS_{2}. Thus, it aqueous layer I

_{2}= 1 − 0.965 = 0.035 g