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4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Intermolecular forces between two benzene molecules are nearly of same strength as those between  two toluene molecules. For a mixture of benzene and toluene, which of the following are not true?

(i) Δmix H= zero

(ii) Δmix V= zero

(iii) These will form minimum boiling azeotrope.

(iv) These will not form ideal solution.
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(iii) and (iv)

Explanation: Intermolecular forces between benzene and toluene molecules in a combination of benzene and toluene molecules would be approximately as strong as those between two benzene molecules and two toluene molecules separately. As a result, the solution will form an ideal solution and follow Raoult's law. As a result, options (iii) and (iv) are false.

New question posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Which of the following factor(s) affect the solubility of a gaseous solute in the fixed volume of liquid solvent?

(a) nature of solute  (b) temperature (c) pressure

(i) (a) and (c) at constant T

(ii) (a) and (b) at constant P

(iii) (b) and (c) only

(iv) (c) only
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New answer posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

KH value for Ar(g), CO2 (g), HCHO (g) and CH4 (g) are 40.39, 1.67, 1.83*10-5 and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

(i) HCHO4

(ii) HCHO2< CH4

(iii) Ar2

(iv) Ar4
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(iii) Ar< CO2< CH4< HCHO

Explanation:  The higher the KH value, the better. The solubility of a gas at a given pressure will be lower, hence the solubility of given gases will increase as KH values rise.

New question posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

KH value for Ar(g), CO2 (g), HCHO (g) and CH4 (g) are 40.39, 1.67, 1.83×10-5 and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

(i) HCHO4

(ii) HCHO2< CH4

(iii) Ar2

(iv) Ar4

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New answer posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

On the basis of information given below mark the correct option.

Information: On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.

(i) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show positive deviation from Raoult's law.

(ii) At  specific composition methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult's law.

(iii) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult's law.

(iv) At specific composition methanol-acetone

...more
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(i) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show  positive deviation from Raoult's law.

Explanation:  (i) (A-A) or (B-B) interactions are more powerful than (A-B) interactions, where A represents a methanol molecule and B represents an acetone molecule. It means that molecules of A (or B) will have an easier time escaping from this solution. This will result in a positive departure from Rault's law when the vapour pressure rises.

(ii) The methanol-acetone mixture forms the smallest boiling azeotrope due to this

...more

New answer posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is _____________.

(i) 0.004

(ii) 0.008

(iii) 0.012

(iv) 0.016
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(iv) 0.016

Explanation: Apply the relation : M1V1= M2V2

Given: M1=0.02M, V1=4L, M2=? V2=5L

Therefore, 0.02*4L=M2*5L

M2=0.08/5

=0.016 M

New answer posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

If two liquids A and B form minimum boiling azeotrope at some specific composition then_______________.

(i) A–B interactions are stronger than those between A–A  or B–B.

(ii) Vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution.

(iii) Vapour pressure of solution decreases because less number of molecules of only one of the liquids escapes from the solution.

(iv) A–B interactions are weaker than those between A–A or B–B.
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(iv) A–B interactions are weaker than those between A–A or B–B.

Explanation: (i) At a given composition, the solutions that demonstrate a big positive divergence from Rault's law form a minimum boiling azeotrope.

(ii) When Rault's law is deviated positively, A-B interactions are weaker than A-A or B-B interactions.

New answer posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Two beakers of capacity 500 mL were taken. One of these beakers, labelled as “A”, was filled with 400 mL water whereas the beaker labelled “B” was filled with 400 mL of 2 M solution of  NaCl. At the same temperature both the beakers were placed in closed containers of same material and same capacity as shown in Fig.2.2. At a given temperature, which of the following statement is correct about the vapour pressure of pure water and that of NaCl solution.

(i) Vapour pressure in container (A) is more than that in container (B).

(ii) Vapour pressure in container (A) is less than that in container (B).

(iii) Vapour pressure i

...more
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(i) Vapour pressure in container (A) is more than that in container (B).

Explanation:  Due to the fleeing inclinations of water molecules from the liquid's surface, the vapour pressure rises. The vapour pressure increases as the number of molecules on the liquid's surface increases. Because only water molecules are present at the surface of beaker A, it has a higher vapour pressure. However, a fraction of the surface area of the solution in beaker B containing NaCl solution is occupied by NaCl molecules, which are non-volatile and have no tendency to escape. As a resul

...more

New answer posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

On the basis of information given below mark the correct option. Information:

(A) In bromoethane and chloroethane mixture intermolecular interactions of A–A and B–B type are nearly same as A–B type interactions.

(B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions.

(C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type  interactions.

(i) Solution (B) and (C) will follow Raoult's law.

(ii) Solution (A) will follow Raoult's law.

(iii) Solution (B) will show negative deviation from  Raoult's law

...more
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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(ii) Solution (A) will follow Raoult's law.

Explanation:  A-A and B-B intermolecular interactions should be almost identical to A-B type interactions in an ideal solution.

New question posted

4 months ago

Solutions Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

We have three aqueous solutions of NaCl labelled as ‘A’,‘B’ and ‘C’ with

Concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor for these solutions will be in the order______.

(i) iA

(ii) iA >iB>iC

(iii) iA=iB=iC

(iv) iAiC
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