CBSE Previous Year Solved Question Papers Class 12 Chemistry 2018. In this blog, the question paper of the year 2018 is being presented to the students of class 12.
CBSE Previous Year Solved Question Papers Class 12 Chemistry 2018.
Q.1 For reaction A –> B, the rate of reaction becomes three times when the
concentration of A is increased by nine times. What is the order of reaction?
Ans:-
The rate of a reaction is typically expressed as:
rate = k[A]^m[B]^n
where k is the rate constant, m, and n are the order of reaction with respect to reactants A and B, respectively, and [A] and [B] are their respective concentrations.
In this case, we know that when the concentration of A is increased by nine times, the rate of reaction becomes three times. Mathematically, this can be expressed as:
(rate 2) = 3(rate 1)
where (rate 2) is the rate of reaction when the concentration of A is increased by nine times, and (rate 1) is the rate of reaction initially.
Using the rate equation above, we can substitute the concentrations and rates into the equation to get:
k([A]2)^m[B]^n = 3k([A]1)^m[B]^n
Simplifying, we get:
([A]2/[A]1)^m = 3
([9[A]1]/[A]1)^m = 3
9^m = 3
m = log(3)/log(9)
m = 0.5
Therefore, the order of reaction with respect to A is 0.5, and the overall order of the reaction is:
m + n = 0.5 + n
We do not have enough information to determine the order of reaction with respect to B.
Q.2 Why is adsorption always exothermic?
Adsorption is the process of adhesion of atoms, ions, or molecules from a gas, liquid or dissolved solid to a surface. This process can be exothermic or endothermic, depending on the nature of the adsorbate and the surface. However, in most cases, adsorption is exothermic, meaning that energy is released during the process.
There are several reasons why adsorption is typically exothermic:
- Formation of intermolecular forces: During adsorption, the adsorbate molecules come into close proximity to the surface and interact with the surface atoms or molecules through intermolecular forces such as Van der Waals forces, hydrogen bonding, or dipole-dipole interactions. These intermolecular forces are attractive and result in the release of energy as heat.
- Decrease in entropy: Adsorption involves the adsorbate molecules becoming confined to a small surface area, which leads to a decrease in the degree of freedom of the adsorbate molecules. This decrease in entropy is associated with a decrease in the overall energy of the system.
- Release of energy due to the formation of chemical bonds: In some cases, the adsorbate molecules may form chemical bonds with the surface atoms or molecules, which results in the release of energy as heat.
Overall, these factors contribute to the exothermic nature of adsorption. However, it is important to note that adsorption can also be endothermic in certain cases, such as when the adsorbate molecules are highly reactive and undergo chemical reactions with the surface.
Q.3 Write the coordination isomer of [Cu(NH3)4 ][PtCl4 ].
Ans:-
The coordination isomers are a type of structural isomer that differ in the way the ligands are arranged around the central metal ion. To find the coordination isomer of [Cu(NH3)4][PtCl4], we need to exchange the positions of the ligands between the two metal ions.
The [Cu(NH3)4] complex has four ammonia (NH3) ligands that are coordinated to the central copper (Cu) ion, while the [PtCl4] complex has four chlorides (Cl) ligands that are coordinated to the central platinum (Pt) ion.
To generate the coordination isomer, we need to exchange one type of ligand between the two complexes. In this case, we can exchange one of the ammonia ligands in [Cu(NH3)4] with one of the chloride ligands in [PtCl4], giving us:
[Cu(NH3)3Cl][PtCl3(NH3)]
In this coordination isomer, the central copper (Cu) ion is coordinated to three ammonia (NH3) ligands and one chloride (Cl) ligand, while the central platinum (Pt) ion is coordinated to three chlorides (Cl) ligand and one ammonia (NH3) ligand.
Q.4 An aromatic organic compound ‘A’ with molecular formula C8H8O gives positive DNP and iodoform tests. It neither reduces Tollens’ reagent nor does it decolorise bromine water. Write the structure of ‘A’
Ans;-
The given information helps us to identify the functional groups present in the given organic compound. The molecular formula of the compound is C8H8O, which suggests that it may contain an aromatic ring. The positive DNP and iodoform tests indicate the presence of a carbonyl group. The compound does not reduce Tollens’ reagent or decolorize bromine water, suggesting the absence of an aldehyde or an alkene group.
Putting all the information together, we can propose that the compound ‘A’ is an aromatic ketone. The structure of the compound can be represented as:
O
||
C6H4
||
C=O
This is the structure of acetophenone, which has the molecular formula C8H8O. It contains an aromatic ring (C6H4) and a carbonyl group (C=O) attached to the ring. Acetophenone gives positive DNP and iodoform tests due to the presence of the carbonyl group, and it does not react with Tollens’ reagent or bromine water because it lacks an aldehyde or an alkene group.
CBSE Previous Year Solved Question Papers Class 12 Chemistry 2018.
Q.5 Predict the major product formed when sodium ethoxide reacts with tert.Butyl chloride.
Ans:-
When sodium ethoxide reacts with tert-butyl chloride, an elimination reaction occurs, leading to the formation of an alkene as the major product.
The reaction mechanism involves the ethoxide ion (C2H5O-) acting as a strong base, abstracting a proton from the tert-butyl chloride (C4H9Cl) to form a tert-butyl carbocation (C4H9+). The carbocation then undergoes the elimination of a chloride ion (Cl-) to give the alkene.
The major product formed is 2-methylpropene (also known as isobutylene), which is an unsymmetrical alkene with the formula C4H8.
The reaction can be represented by the following equation:
C4H9Cl + C2H5O- → C4H8 + NaCl + C2H6O
where NaCl is formed as a byproduct and C2H6O is ethanol, which is produced as a solvent in this reaction.
Q. 6 Why does a mixture of Carbon disulfide and acetone show positive deviation from
Raoult’s law? What type of azeotrope is formed by this mixture?
Ans:-
A mixture of carbon disulfide (CS2) and acetone (CH3COCH3) shows positive deviation from Raoult’s law because the intermolecular forces between unlike molecules are stronger than those between like molecules. In other words, the attractive forces between CS2 and CH3COCH3 molecules are stronger than the attractive forces between CS2-CS2 or CH3COCH3-CH3COCH3 molecules. This leads to a higher vapor pressure of the mixture than what would be predicted by Raoult’s law.
The positive deviation from Raoult’s law is due to the fact that CS2 and CH3COCH3 are chemically different, and their molecules experience different types of intermolecular forces. Specifically, CS2 has a linear molecular geometry and exhibits dipole-dipole interactions and London dispersion forces, while CH3COCH3 has a non-linear molecular geometry and exhibits hydrogen bonding, dipole-dipole interactions, and London dispersion forces.
When a mixture of CS2 and CH3COCH3 is heated, the boiling point of the mixture decreases, and a constant boiling mixture, also known as an azeotrope, is formed. This azeotrope has a boiling point lower than that of either pure component, indicating that the mixture is more volatile than either component alone. The azeotrope formed by this mixture is a minimum boiling azeotrope, which means that the composition of the vapor in equilibrium with the liquid mixture has the same composition as the liquid mixture itself. In other words, the azeotrope is a constant boiling mixture that contains a fixed ratio of CS2 and CH3COCH3, and this ratio cannot be changed by further distillation. The composition of the azeotrope is approximately 62% CS2 and 38% CH3COCH3 by weight.
Q.7 Which one of the following compounds is more reactive towards SN 2 reaction and
why?
CH3CH(Cl)CH2CH3 or CH3CH2CH2Cl
Ans:-
CH3CH2CH2Cl is more reactive towards SN2 (Substitution Nucleophilic Bimolecular) reaction compared to CH3CH(Cl)CH2CH3 because of the following reasons:
In the SN2 reaction, the nucleophile attacks the carbon center that is bonded to the leaving group, and the leaving group departs simultaneously. The rate of the SN2 reaction depends on the steric hindrance at the carbon center, as well as the ability of the nucleophile to approach the carbon center.
In the case of CH3CH2CH2Cl, the carbon center attached to the chlorine atom is a primary carbon, which has less steric hindrance than the secondary carbon in CH3CH(Cl)CH2CH3. This means that the nucleophile can more easily approach the carbon center in CH3CH2CH2Cl compared to CH3CH(Cl)CH2CH3.
Additionally, the carbon-chlorine bond in CH3CH2CH2Cl is longer and weaker than the carbon-chlorine bond in CH3CH(Cl)CH2CH3 due to the difference in the hybridization of the carbon atoms. The carbon in CH3CH2CH2Cl is sp3 hybridized, while the carbon in CH3CH(Cl)CH2CH3 is sp2 hybridized. The weaker carbon-chlorine bond in CH3CH2CH2Cl is more easily broken, making it a better-leaving group, which also increases the reactivity towards the SN2 reaction.
Therefore, CH3CH2CH2Cl is more reactive towards SN2 reaction compared to CH3CH(Cl)CH2CH3.
FOR BSC
Q.8 A current of 1.50 A was passed through an electrolytic cell containing AgNO3
solution with inert electrodes. The weight of the silver deposited was 1.50 g. How long
did the current flow? (Molar mass of Ag = 108 g mol–1, 1F = 96500 C mol–1)
Ans:-
The amount of substance produced or consumed in an electrolytic cell is directly proportional to the quantity of electricity passed through the cell. The relationship between the amount of substance produced and the quantity of electricity passed is given by Faraday’s laws of electrolysis.
The first law of electrolysis states that the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolytic cell.
The quantity of electricity passed through the electrolytic cell is given by:
Quantity of electricity = Current x Time
In this problem, the weight of silver deposited is given as 1.50 g, and the molar mass of silver is 108 g mol–1. Therefore, the amount of silver produced can be calculated as:
Amount of Ag = 1.50 g / 108 g mol-1 = 0.0139 mol
The quantity of electricity passed through the electrolytic cell can be calculated as:
Quantity of electricity = (Amount of Ag) x (1F/ 1 mol of e-) = (0.0139 mol) x (96500 C mol-1) = 1341.5 C
The current passing through the cell is given as 1.50 A. Therefore, the time taken for the current to flow through the cell can be calculated as:
Time = (Quantity of electricity) / (Current) = 1341.5 C / 1.50 A = 894 seconds
Therefore, the current flowed for 894 seconds or 14.9 minutes (approx).
CBSE Previous Year Solved Question Papers Class 12 Chemistry 2018.
or
The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 10–4 S cm–1 .
Calculate the molar conductivity (m ) of the solution.
Ans:-
The molar conductivity (𝜇m) of a solution is defined as the conductivity of the solution divided by the molar concentration of the electrolyte.
𝜇m = 𝜎 / c
where 𝜎 is the conductivity of the solution, and c is the concentration of the solution in mol L^-1.
In this problem, the concentration of acetic acid solution is given as 0.01 M, and the conductivity of the solution is given as 1.65 × 10^-4 S cm^-1.
Therefore, the molar conductivity of the solution can be calculated as:
𝜇m = 𝜎 / c = (1.65 × 10^-4 S cm^-1) / (0.01 mol L^-1)
𝜇m = 0.0165 / 0.01
𝜇m = 1.65 S cm^2 mol^-1
Therefore, the molar conductivity of the 0.01 M acetic acid solution at 298 K is 1.65 S cm^2 mol^-1.
Q.9Identify the following :
(i) Transition metal of 3d series that exhibits the maximum number of oxidation
states.
(i) The transition metal of the 3d series that exhibits the maximum number of oxidation states is Manganese (Mn). Mn can exhibit oxidation states ranging from -3 to +7, with +2, +4, and +7 being the most common.
(ii) An alloy consisting of approximately 95% lanthanoid metal used to produce
bullet, shell, and lighter flint
(ii) The alloy consisting of approximately 95% lanthanoid metal used to produce bullet, shell, and lighter flint is Mischmetal. It is a mixture of various rare earth metals, with the most common composition being about 50% cerium, 25% lanthanum, and smaller amounts of other rare earth metals. Mischmetal is used to produce sparks for lighters and to create sparks in flintlock firearms. It is also used in the production of alloys used in the manufacture of magnets and batteries.
Q.10 Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of Magnesium bromide = 184 g mol–1, Kf for water = 1.86 K kg mol–1)
When a solute is added to a solvent, the freezing point of the resulting solution is lowered. The extent of this lowering is given by the equation:
ΔTf = Kf x molality
where ΔTf is the freezing point depression, Kf is the freezing point depression constant for the solvent (water in this case), and molality is the molal concentration of the solution, which is defined as the number of moles of solute per kilogram of solvent.
To calculate the molality of the solution, we first need to calculate the number of moles of Magnesium bromide in 10.5 g of the compound:
Number of moles of MgBr2 = Mass / Molar mass = 10.5 g / 184 g mol^-1 = 0.057 mole
Next, we need to calculate the mass of water in the solution:
Mass of water = 200 g
Now, we can calculate the molality of the solution:
Molality = Number of moles of solute / Mass of solvent in kg
Molality = 0.057 mole / (200 g / 1000) kg = 0.285 mol kg^-1
Finally, we can use the equation for freezing point depression to calculate the freezing point of the solution:
ΔTf = Kf x molality
ΔTf = (1.86 K kg mol^-1) x (0.285 mol kg^-1) = 0.531 K
The freezing point of pure water is 0°C or 273.15 K. Therefore, the freezing point of the solution will be:
The freezing point of solution = Freezing point of solvent – ΔTf
Freezing point of solution = 273.15 K – 0.531 K = 272.619 K
Therefore, the freezing point of the aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide, is 272.619 K.
