CUET UG Chemistry: Mock Test - 2 Free Online Test 2026

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.

Given, for NaCl, HCl and NaOAc are 126.4, 425.9 and 91.0 S cm² mol^-1

= (425.9 + 91.0 – 126.4) S cm² mol^–1

= 390.5 S cm2 mol–1

For redox change

n-factor = 5

n_eq = nmol × n-factor

= 1 × 5

= 5

Charge required in Faraday = Number of n_eq

= 5 F

On electrolysis of aq. NaCl

Cathode:

Anode:

Net reaction :

K = 2.3 × 10^-5 L mol^-1 s^-1

For order of reaction,

The general formula for the unit of k is mol^1-n L ^n-1 s^-1

The unit of rate constant

The unit ‘L mol^–1 s^–1 ’ corresponds to the 2nd-order reaction.

• Option A: True. Actinoids are highly reactive, especially when finely divided, due to their metallic nature.
• Option B: True. Actinoid contraction is greater than lanthanoid contraction because 5f electrons have poorer shielding than 4f electrons, leading to a stronger nuclear pull.
• Option C: False. Most trivalent lanthanoid ions (e.g., La³⁺, Ce³⁺, Gd³⁺, Lu³⁺) are colorless in both solid and aqueous states due to the absence of f-f transitions (e.g., f⁰, f⁷, f¹⁴ configurations). Only some lanthanoids with partially filled f-orbitals (e.g., Nd³⁺, Er³⁺) show color.
• Option D: True. Lanthanoids have metallic structures, making them good conductors of heat and electricity.

Thus, the correct answer is C.

The correct answer is Weak electrolytes do not dissociate completely.

• Colligative properties depend on the number of solute particles in a solution, not the type of particles. Weak electrolytes, such as acetic acid, do not dissociate completely into ions when dissolved in a solvent.
• This incomplete dissociation means that the total number of particles added to the solution is relatively low compared to strong electrolytes, which dissociate fully.
• Therefore, the addition of a small amount of a weak electrolyte has a less significant effect on the solution's colligative properties because it contributes fewer particles to the solution than a strong electrolyte would.

So, the colligative properties not significantly change with the addition of a small amount of a weak electrolyte to a solution since Weak electrolytes do not dissociate completely.

The correct answer is They are both linearly proportional

• Both boiling point elevation and freezing point depression are linearly proportional to the molal concentration of the solute in the solution.
• This relationship is described by formulas (ΔTb=Kb⋅mΔTb=Kb⋅m) for boiling point elevation and (ΔTf=Kf⋅mΔTf=Kf⋅m) for freezing point depression, where (ΔTb) and (ΔTf) are the changes in boiling and freezing points, respectively, (Kb) and (Kf) are the ebullioscopic and cryoscopic constants, and (m) is the molality of the solution.
• This linear proportionality makes it easier to predict the changes in these properties based on the concentration of the solute.

So, the correct answer is They are both linearly proportional. 

Freezing point depression is a colligative property where a solute lowers the freezing point of a solvent. Adding salt (e.g., NaCl) to water on roads dissolves into ions, reducing the freezing point of the resulting solution below 0°C. This prevents ice formation, enhancing road safety in winter.

Correct answer: C: To decrease the freezing point of water.

According to Raoult’s law, adding a non-volatile solute to a solvent reduces the solvent’s vapor pressure. The solute particles occupy the solution’s surface, decreasing the number of solvent molecules that can escape into the vapor phase.

Correct answer: B: It decreases the vapor pressure.

Osmotic pressure (π \pi π) is a colligative property that measures the pressure needed to prevent solvent flow across a semipermeable membrane from a dilute to a concentrated solution. It is given by:
π = iCRT
Where π is the osmotic pressure, i is the van’t Hoff factor, M is the molar concentration, R is the universal gas constant and T is the temperature. Osmotic pressure is directly proportional to solute concentration, reflecting the presence of solute particles.

Correct answer: D: The presence of solute particles and is proportional to solute concentration.

2-bromobutane → CH₃CH = CHCH₃ + CH₃CH₂CH=CH₂

The major product follows Saytzeff rule.

Hence option (A) is the correct answer.

 Because it is Open chain compound /Aliphatic Compound and not aryl halide as it does not contain halogen atom. It's simply  alkane.

 ΔT_f = 0 - ( - 0.00732° ) = 0.00732°
ΔT_f = Molality x K_f x i

0.00732 = 0.002 x 1.86 x i

i = 1.97 = 2,0

Diphenhydramine​ is an antihistamine mainly used to treat allergies. It can also be used for insomnia, symptoms of the common cold, tremor in parkinsonism, and nausea. It is used by mouth, injection into a vein, injection into a muscle, or applied to the skin.

Nitro compounds can be reduced using iron scrap and hydrochloric acid, resulting in the formation of amines.

The reaction can be summarised as follows:

• RNO₂ + Fe/HCl → RNH₂

This process effectively converts nitro compounds into amines. The presence of iron and hydrochloric acid facilitates the reduction reaction, leading to the desired product.

Ephedrine is a medication that is commonly used to increase blood pressure. Here are some key points about its usage:

• Ephedrine acts as a stimulant, promoting the release of certain hormones.
• It is often used in medical settings to manage low blood pressure.
• This medication can also be used to treat certain types of asthma.

The correct answer is Ribose.

• Ribose is a type of pentose monosaccharide.

• Pentoses are sugars that contain five carbon atoms.

• Other options, like Glucose, Fructose, and Galactose, are hexoses, which have six carbon atoms.

The ring structure of glucose is formed through the creation of a hemiacetal.

This occurs when the hydroxyl group on carbon 5 reacts with the aldehyde group on carbon 1.

Key points about this reaction include:

• The reaction leads to the formation of a ring structure, specifically a six-membered ring known as a pyranose.
• The conversion to a hemiacetal is crucial for the stability and functionality of glucose.
• This structural change influences how glucose interacts with other molecules in biological systems.

Glucose is a type of sugar known as a monosaccharide.

A monosaccharide is the simplest form of carbohydrate, consisting of a single sugar unit. Here are some key points about glucose:

• Basic unit: It is the most basic form of carbohydrate.
• Energy source: Glucose serves as a primary energy source for living organisms.
• Structural role: It can also play a role in forming larger carbohydrates.

The inner transition elements are the elements in which the added electrons go to (n-2) f-orbitals. These include lanthanoids and actinoids. In lanthanoids, the added electron goes to 4f orbital and in actinoids, the added electron goes to 5f orbital.

Let the oxidation state of pt be x.
Oxidation state of cl is -1.
So x + 0 – (1*3) must be equal to -1 since the charge of the whole compound is -1.
x + 0 – (1*3) = -1
x -3 = -1
x = -1 + 3
x = +2
So the oxidation state of platinum is +2.
Secondary is due to legend there are mono deadened legend 
then 1×4= 4
 

Pt(NH₃)₂CI₄] is neutral complex and no ions are formed in aqueous solution.

K,(PtCl₆] → 2K⁺ + [PtCI₆]^2-
It does not have Cl^- ion toffive precipitate of AgCI.

tert-Butyl alcohol is the most heavily branched isomer and is tightly packed compared to the others. This decreases the effective surface area of the alkyl part which is hydrophobic in nature.

Phenol’s -OH group activates the aromatic ring, making it highly reactive toward electrophilic aromatic substitution like bromination. Unlike benzene, which requires Lewis acids (e.g., FeBr₃) to generate Br⁺, phenol reacts directly with Br₂ in a non-polar solvent like CS₂, which controls the reaction and favors monobromination (e.g., 4-bromophenol).
- B, C, D: FeBr₃, AlCl₃, and BF₃ are Lewis acids needed for less reactive aromatics like benzene, not phenol.
Correct option: A

Phenol (C₆H₅OH) has limited solubility in water (~8.3 g/100 mL) due to the large non-polar aryl group, which outweighs the polar -OH group’s hydrogen bonding. It is more soluble in:
- B: Alcohol, due to similar polarity and hydrogen bonding.
- C: Ether, due to compatibility with the aryl group.
- D: NaOH, where phenol reacts to form soluble sodium phenoxide (C₆H₅ONa).
Water is the least soluble medium.
Correct option: A

Four compounds (A, B, C, D) with similar molecular masses (~58–60 g/mol) are compared. Alcohols have the highest boiling points due to intermolecular hydrogen bonding, absent in hydrocarbons, haloalkanes, and ethers. Example:
- A: Hydrocarbon (e.g., butane, 58 g/mol, -0.5°C).
- B: Haloalkane (e.g., 1-chlorobutane, 78.5 g/mol, 78°C).
- C: Alcohol (e.g., 1-propanol, 60 g/mol, 97°C).
- D: Ether (e.g., diethyl ether, 74 g/mol, 34.6°C).
Compound C, an alcohol, has the highest boiling point.
Correct option: C

Molarity is defined as the number of moles of solute dissolved per litre of solution (IV). Molality refers to moles of solute per kilogram of solvent (I), making it temperature-independent. Mole fraction is the ratio of moles of one component to the total moles of all components in the solution (II). Mass percentage expresses the mass of solute per 100 grams of solution (III), also called weight percentage.

Relative lowering of vapour pressure occurs because solute particles reduce the escaping tendency of solvent molecules, and it decreases the vapour pressure (IV). Elevation of boiling point increases as more solute is added, because the vapour pressure must be increased to reach the boiling point (III). All colligative properties depend on the number of solute particles (I), but depression of freezing point is most directly associated with this principle in context. Osmotic pressure is the principle behind reverse osmosis, widely used in water purification by applying pressure greater than osmotic pressure (II).

ΔTᵇ = i × Kᵇ × m, so the elevation of boiling point depends on the van’t Hoff factor i. Urea is a non-electrolyte (i = 1); KCl gives 2 ions (i = 2); BaCl₂ gives 3 ions (i = 3); K₃[Fe(CN)₆] gives 4 ions (i = 4). Therefore, in increasing order of ΔTᵇ: Urea < KCl < BaCl₂ < K₃[Fe(CN)₆], i.e., (B) < (A) < (C) < (D).

The standard reduction potentials (E°) are: Ag⁺/Ag = +0.80 V, Cu²⁺/Cu = +0.34 V, Fe²⁺/Fe = −0.44 V, and Zn²⁺/Zn = −0.76 V. A higher (more positive) E° indicates a stronger tendency to get reduced and a weaker reducing agent. Therefore, the decreasing order of standard reduction potential is: Ag > Cu > Fe > Zn, i.e., (D) > (C) > (B) > (A).