SRMJEEE Subject Wise & Full Length Chemistry Mock Test - 7 Free Online

The major component of dynamite is nitroglycerin.

Nitroglycerin is a highly explosive liquid that is used in the manufacture of dynamite. It is known for its sensitivity and power. Here are some key points about it:

• Composition: Nitroglycerin is derived from glycerol and nitric acid.
• Stability: It is more stable when mixed with an absorbent material, which is how it forms dynamite.
• Usage: Besides dynamite, nitroglycerin is also used in medical applications, such as treating heart conditions.

To convert RCO to RCH₂, the following reactions can be considered:

• Clemensen's reduction is effective as it reduces carbonyl compounds to alkanes.
• Wolf-Kishner reduction is also suitable for converting carbonyl groups to methylene groups.
• HI/red P at 200°C can facilitate this transformation by reducing carbonyl compounds.
• Wurtz reaction is not applicable here as it typically couples alkyl halides rather than reducing carbonyl compounds.

Therefore, the Wurtz reaction is the one that cannot be used for this conversion.

RNA differs from DNA primarily due to its composition.

• RNA contains ribose sugar instead of deoxyribose, which is found in DNA.
• In RNA, the base uracil replaces thymine, which is present in DNA.

These differences contribute to the distinct roles of RNA and DNA in biological systems.

To determine which compound has the lowest boiling point, consider the following factors:

• Isobutane: A branched alkane with a lower boiling point due to reduced surface area.
• I-butyne: An alkyne that typically has a higher boiling point due to stronger intermolecular forces.
• I-butene: An alkene with a moderate boiling point, influenced by its double bond.
• n-butane: A straight-chain alkane with a higher boiling point compared to isobutane.

The compound with the minimum boiling point is likely to be isobutane, as its branched structure leads to lower boiling due to weaker van der Waals forces compared to the other compounds listed.

The particle with the same mass as an electron is the positron.

Key points about the positron include:

• The positron is the antimatter counterpart of the electron.
• Both particles have the same mass, approximately 9.11 x 10^-31 kg.
• While they have equal mass, they possess opposite charges; the electron is negatively charged, and the positron is positively charged.

In contrast:

• The proton has a much greater mass, about 1836 times that of an electron.
• The neutron is also significantly heavier than an electron.
• A photon is massless and does not have mass like an electron.

The intensity of the brown colour of bromine vapours will remain largely unchanged when adding certain substances. Here’s a breakdown of the options:

• Pieces of marble: These are mainly composed of calcium carbonate and will not interact significantly with bromine vapours, thus having minimal effect on colour intensity.
• Animal charcoal powder: This substance is known for its adsorptive properties and may absorb some bromine, potentially altering the colour.
• Carbon tetrachloride: This solvent is non-reactive with bromine and will not diminish the colour intensity.
• Carbon disulphide: Similar to carbon tetrachloride, this liquid will not react with bromine, preserving the vapour’s colour.

In conclusion, the addition of either carbon tetrachloride or carbon disulphide will ensure that the intensity of the brown bromine vapours does not decrease appreciably.

Benzene is primarily obtained through fractional distillation of:

• Light oil - This fraction contains lighter hydrocarbons, making it a common source of benzene.
• Middle oil - It can also yield benzene, but to a lesser extent than light oil.
• Anthracene oil - This is a less common source, primarily used for other aromatic compounds.
• Heavy oil - Typically not a primary source for benzene due to its heavier fractions.

Of these options, light oil is the most significant source for obtaining benzene through fractional distillation.

To determine the maximum amount of BaCl₄ that precipitates when mixing BaCl₂ with H₂SO₄, consider the following:

• BaCl₂ has a concentration of 0.5 M.
• H₂SO₄ has a concentration of 1 M.
• The reaction produces BaSO₄, which is the precipitate.
• Each mole of BaCl₂ reacts with one mole of H₂SO₄ to produce one mole of BaSO₄.
• Thus, the limiting reagent will dictate the amount of precipitation.
• Since BaCl₂ is present in a lower concentration (0.5 M), this will determine the maximum amount of BaSO₄ formed.

The maximum amount of BaCl₄ precipitation corresponds to the concentration of BaCl₂, which is 0.5 M.

The energy of a C-C triple bond in acetylene is approximately:

• 140 k.cals per mole.
• This value reflects the strength of the bond, making it one of the strongest types of carbon bonds.
• The energy can vary slightly depending on the source but typically falls within the range of 120 to 200 k.cals.

Lower carboxylic acids are soluble in water due to:

• Hydrogen bonding: These acids can form strong interactions with water molecules, enhancing solubility.
• Dissociation into ions: Carboxylic acids can ionise in water, contributing to their solubility.

While lower molecular mass can aid in solubility, the primary reasons are hydrogen bonding and ionisation.

The weakest type of bond among the following is the van der Waals forces.

• Ionic bonds are strong interactions formed between charged ions.
• Metallic bonds involve a lattice of metal atoms sharing electrons, providing strength and conductivity.
• Covalent bonds are formed when atoms share electrons, resulting in strong connections.
• Van der Waals forces are much weaker intermolecular forces, arising from temporary dipoles.

This hierarchy indicates that van der Waals forces have significantly less strength compared to the other types of bonds.

The combustion of methane is represonted as
CH₄ + 2O₂ → CO₂ + 2H₂O
Δn = 1 - 3 = - 2
∴ ΔH - ΔU = - 2RT = - 2 x 8.314 x 300

The chemical equilibrium of a reversible reaction is not affected by:

• Catalysts: These speed up the reaction but do not change the position of equilibrium.
• Pressure: Only relevant for reactions involving gases and can shift equilibrium if the number of gas molecules differs on either side.
• Temperature: Changes can shift equilibrium, depending on whether the reaction is exothermic or endothermic.
• Concentration: Altering the concentration of reactants or products will shift the equilibrium to restore balance.

When the initial concentration is doubled, the half-life of the reaction also doubles. This observation provides insights into the order of the reaction:

• Zero order: The half-life is independent of concentration. Doubling concentration does not affect the half-life.

• First order: The half-life remains constant regardless of concentration changes.

• Second order: The half-life increases as concentration decreases. Doubling concentration would halve the half-life.

• Third order: The relationship is more complex but generally does not double the half-life with a concentration change.

Given that doubling the concentration also doubles the half-life, the reaction does not fit the common patterns of zero, first, or second order. Thus, the order is considered undefined.

To determine the enthalpy of formation of carbon monoxide (CO):

• The first reaction shows the formation of carbon dioxide (CO₂) from carbon (C) and oxygen (O₂):
  • C + O₂ → CO₂ ; ∆H° = −x kJ
• The second reaction indicates the formation of CO₂ from carbon monoxide (CO):
  • 2CO + O₂ → 2CO₂ ; ∆H° = −y kJ
• To find the enthalpy of formation of CO, we need to manipulate these reactions:
• By reversing the first reaction, we get:
  • CO₂ → C + O₂ ; ∆H° = +x kJ
• Then, we can write the formation of CO from C and O₂:
  • C + ½O₂ → CO ; ∆H° = ?
• Combining these reactions gives:
  • 2CO → CO₂ + CO₂ ; ∆H° = -y kJ
  • Thus, the overall equation is:
    • 2(C + ½O₂ → CO) = y - 2x
• From this, we deduce that the enthalpy of formation of CO is:
  • y - 2x

Paracetamol serves crucial roles in medical treatment:

• Analgesic: Provides relief from pain.
• Antipyretic: Reduces fever.

However, it is not effective as an antimalarial agent.

CO₂ and N₂O are isoelectronic with 22 electrons and isostructural with linear geometry.
Structure of nitrous oxide:

Structure of carbon di-oxide:

O = C = O

The amount of Aluminium deposited when 0.1 Faraday current is passed through aluminium chloride will be (R = 27)

To find the mass of Aluminium deposited, follow these steps:

• Determine the number of moles of Aluminium deposited using Faraday's laws of electrolysis.
• One Faraday corresponds to the charge needed to deposit one mole of Aluminium (Al).
• Since Aluminium has a valency of 3, it requires 3 Faradays to deposit 1 mole of Al.
• Calculate the moles of Aluminium deposited with 0.1 Faraday:
• Moles of Al = 0.1 Faraday / 3 = 0.0333 moles
• Now, calculate the mass using the formula:
• Mass = Moles × Molar mass
• Mass = 0.0333 moles × 27 g/mole = 0.9 g

Thus, the amount of Aluminium deposited is 0.9 g.

The gas constant R in the ideal gas equation has a specific dimension that is essential for calculations involving gases.

• The dimension of R is expressed as litre-atm/degree K/mole.
• This means it relates the volume of gas, pressure, temperature, and the amount of substance (in moles).
• Understanding R's dimension helps in applying the ideal gas law correctly in various situations.

To determine the empirical formula of the hydrocarbon from the combustion products:

• Mass of CO₂ produced: 12.571 g
• Mass of water produced: 5.143 g

Next, we calculate the moles of each product:

• Moles of CO₂: 12.571 g / 44.01 g/mol = 0.285 moles
• Moles of H₂O: 5.143 g / 18.02 g/mol = 0.286 moles

From the moles of CO₂ and H₂O, we can find the moles of carbon and hydrogen:

• Carbon: Each mole of CO₂ contains 1 mole of C, so there are 0.285 moles of C.
• Hydrogen: Each mole of H₂O contains 2 moles of H, so there are 0.572 moles of H (0.286 moles x 2).

Next, we find the simplest ratio of C to H:

• Divide both by the smallest number of moles (0.285):
• C: 0.285 / 0.285 = 1
• H: 0.572 / 0.285 ≈ 2

The empirical formula of the hydrocarbon is thus CH₂.

A solution of NH₄Cl is

• NH₄Cl (ammonium chloride) is a salt formed from a strong acid (HCl) and a weak base (NH₃).
• When dissolved in water, it dissociates into NH₄⁺ and Cl⁻ ions.
• The NH₄⁺ ion can donate protons (H⁺) to water, producing hydronium ions (H₃O⁺).
• This increase in hydronium ions results in an acidic solution.

Metamerism refers to the phenomenon where compounds with the same molecular formula have different structural arrangements, particularly in the context of positional isomerism. To determine which compounds exhibit metamerism, we can analyse each option by considering their structures and substituents.

• CH₃COC₃H₇: This compound can show metamerism because it has a carbon chain that can be arranged differently while maintaining the same molecular formula.
• CH₃SC₂H₅: This compound does not exhibit metamerism as the substituents are fixed and cannot rearrange to form different structures.
• CH₃ - O - CH₃: As a symmetrical compound (dimethyl ether), it does not show metamerism due to its identical substituents.
• CH₃ - O - C₂H₅: This compound can exhibit metamerism since it contains different substituents that can be arranged in various ways.

In conclusion, the compounds that show metamerism are CH₃COC₃H₇ and CH₃ - O - C₂H₅.

To determine the half-life of the radioactive isotope, follow these steps:

• After 192 minutes, only 1/16 of the original amount remains.
• This indicates that the isotope has undergone four half-lives because:
• 1/2 (after 1 half-life)
• 1/4 (after 2 half-lives)
• 1/8 (after 3 half-lives)
• 1/16 (after 4 half-lives)
• To find the half-life, divide the total time by the number of half-lives:
• 192 minutes ÷ 4 = 48 minutes

The half-life of the radioactive isotope is therefore 48 minutes.

(NH₄)₂Sx reduces one - NO₂ to - NH₂ group

The iodoform test is used to detect certain types of compounds, particularly those containing a methyl group adjacent to a carbonyl group or alcohols that can be oxidised to such compounds.

• Acetophenone - This compound contains a methyl group next to a carbonyl, so it gives a positive iodoform test.
• Ethanal - As an aldehyde with a methyl group, it also gives a positive result.
• Benzophenone - This compound does not have a methyl group next to a carbonyl, hence it will not produce a positive iodoform test.
• Ethanol - Although it can be oxidised to ethanal, it will not directly give a positive iodoform test.

Therefore, the compound that does not give the iodoform test is Benzophenone.

To separate compound C from the mixture, the following method can be employed:

• Distillation is effective for separating components based on their boiling points. Since C transitions from solid to vapour when heated, it can be collected as it evaporates.

This method allows for the efficient separation of C while leaving compounds A, B, and D behind in solid form.

To determine the oxidation number of carbon in C₆H₁₂O₆:

• Carbon (C) can have various oxidation states depending on its bonding environment.
• In glucose (C₆H₁₂O₆), we need to consider the overall structure and the other elements present.
• Hydrogen (H) typically has an oxidation number of +1.
• Oxygen (O) usually has an oxidation number of -2.

The overall charge of glucose is neutral, so we can set up the equation:

• 6x + 12(+1) + 6(-2) = 0, where x is the oxidation number of carbon.
• This simplifies to: 6x + 12 - 12 = 0.
• Thus, 6x = 0, leading to x = 0.

Therefore, the oxidation number of carbon in C₆H₁₂O₆ is 0.