Chemistry Test

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Chemistry Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

The order of acidity of the following compounds is

A
IV > II > III > I

B
IV > II > I > III

C
I > II > III > IV

D
II > IV > III > I

Solution

Presence of OH groups, ortho to COOH, increases the acidity of the compound. The main reason being the stabilisation of carboxylate ion through hydrogen bonding with OH group. Presence of OH group at p-position reduces acidity. Thus, the correct order should be IV > II > I > III.

This is the correct order.
Question 2
1 / -0

The drift speed of Bovine Serum Albumin (BSA) under the influence of electric field in aqueous solution was monitored at several values of pH and the speed values are listed below (opposite signs indicates opposite direction of travel). What is the isoelectric point of the protein?

A
pH = 0

B
pH = 4.8

C
pH = 5.8

D
pH = 6.8

Solution

The data is plotted in the figure below. The drift speed (electrophoretic mobility) is zero at pH = 4.8.

Hence, the isoelectric point is pH = 4.8
Question 3
1 / -0

During winter, the moisture condenses in the form of dew and can be seen on plant leaves and grass. The entropy of the system, in such cases, decreases as liquids possess lesser disorder as compared to gases. With reference to the second law, which of the following statements is correct for the above process?

A
The randomness of the universe decreases.

B
The randomness of the surroundings decreases.

C
Increase in randomness of surroundings is equal to the decrease in randomness of system.

D
The increase in randomness of the surroundings is greater as compared to the decrease in randomness of the system.

Solution

As dew formation is a spontaneous process, entropy or randomness of the universe will increase.
Question 4
1 / -0

Among cellulose, poly(vinylchloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is the weakest is

A
nylon

B
poly vinyl chloride

C
cellulose

D
natural rubber

Solution

Cellulose and nylon are fibres and their intermolecular forces are the strongest. Poly vinyl chloride is a thermoplastic, the intermolecular force in it is neither strong nor weak.

Natural rubber is an elastomer and the polymers are held by weak van der Waals forces.

Hence, natural rubber has the weakest forces.
Question 5
1 / -0

Which of the following is not a buffer solution?

A
0.8 M H₂S + 0.8 M KHS

B
2 M C₆H₅NH₂ + 2 M C₆H₅NH₃⁺Br^-

C
3 M H₂CO₃ + 3 M KHCO₃

D
0.05 M KClO₄ + 0.05 M HClO₄

Solution

A buffer solution is that which can resist a change in pH on addition of small amount of acid or base.

It is of two types:

(i) An acidic buffer (weak acid in excess + salt of its conjugate base)

(ii) A basic buffer (weak base in excess + salt of its conjugate acid)

KClO₄ + HClO₄ is not a buffer solution because it a mixture of a strong acid (HClO₄) and the salt of its conjugate base (KClO₄).

Note: All the other mixtures in the options are perfect buffers.
Question 6
1 / -0

In the reaction P + Q -> R + S, the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is

A
2

B
1

C
0

D
3

Solution

For P, if t_50% = x, then t_75% = 2x

Hence, the order of the reaction w.r.t P is 1.

Q decreases linearly with time. So, rate with respect to Q remains constant. Hence, order of reaction with respect to Q is zero.

The rate law expression,

r = k[P]¹[Q]⁰

So, overall order is 1 + 0 = 1
Question 7
1 / -0

Which of the following statements are true?

(A) Sucrose is a non-reducing sugar.

(B) Glucose is oxidised by bromine water.

(C) Glucose rotates plane polarised light in clockwise direction.

(D) Fructose is oxidised by bromine water.

A
(A), (B) and (C)

B
(A) and (B)

C
(B) and (C)

D
(A) and (D)

Solution

The statement (D) is incorrect. Fructose is a ketohexoses and is not oxidised by mild oxidising agents, like chlorine or bromine water.

All the other statements are correct.

(A) Sucrose has an anomeric carbon which is not free. The carbon that links glucose and fructose and fructose does not have free OH group to undergo reducing reaction and to open the ring, sucrose is non-reducing.

(B) With bromine water, glucose is oxidised to gluconic acid.

(C) Glucose is dextrorotatory and rotates the plane polarised light in the clockwise direction.
Question 8
1 / -0

Calculation the boiling point of a 1m aqueous solution (density \(1.04 \mathrm{~g} \mathrm{~mL}^{-1}\) ) of potassium chloride (Kb for water \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1},\) Atomic masses: \(\mathrm{K}=39 \mathrm{u},\) CI=39.9u)Assume, potassium chloride is completely dissociated in solution.

A
107.281^oC

B
103.680^oC

C
101.078^oC

D
None of these

Solution

Volume of solution = 1000 ml

Mass of the solution = V × d = 1000 mL × 1.04 g/mL = 1040 g

Amount of solute in 1000 mL solution = 1

M = 1 M × Molecular mass = 1 M × 74.5 g/mol = 74.5 g

Mass of water = Mass of solution - Mass of KCl = 1040 g - 74.5 g = 965.5 g

Molality of the solution = m

= 1.078^oC

Boiling point of the solution = 100^oC + 1.078^oC

= 101.078^oC
Question 9
1 / -0

This compound, on hydrolysis in aqueous acetone, will give:

A
mixture of (i) and (ii)

B
mixture of (i) and (iii){{}}

C
only (iii)

D
only (i)

Solution

It is SN¹ reaction in which an intermediate carbocation is involved.

This carbocation is specially stabilised through resonance in which -O-CH₃ group acts as a good electron donor, hence we will get a mixture of two products.
Question 10
1 / -0

During the separation of Group IV metal ions as insoluble carbonates by adding a saturated solution of (NH₄)₂CO₃ to the solution of the sample in an ammoniacal medium containing excess of NH4Cl, Mg²⁺ is not precipitated either as Mg(OH)₂ or as MgCO₃ because

A
the concentrations of OH^- and CO₃^- ions are low due to common ion effect that the ionic product of Mg(OH)₂ and MgCO₃ is zero

B
solubility product values of Mg(OH)₂ and MgCO₃ decrease due to common ion effect

C
solubility product values of Mg(OH)₂ and MgCO₃ increase due to common ion effect

D
the precipitates of Mg(OH)₂ and MgCO₃, initially formed, combine to form a soluble carbonate i.e. 4MgCO₃.Mg(OH)₂.

Solution

Since NH₃ is a weak base, its dissociation is suppressed due to the presence of the common ion NH₄⁺ in high concentration from NH₄Cl causing the concentration of OH- to be very low. Due to the same reason, the concentration of CO₃^2- is also very low. So, Mg(OH)₂ and MgCO₃ are not precipitated as their solubility products and are not exceeded by the corresponding ionic products. Hence, this statement is correct.