MHT CET 2024 Chemistry Question Paper with Answer and Solution

(B) The energy $E$ of a sound wave is proportional to the square of its frequency $n$ and the square of its amplitude $A$.
$E \propto n^2 A^2$.
Given that the energies of the waves produced by $P$ and $Q$ are equal,we have:
$n_P^2 A_P^2 = n_Q^2 A_Q^2$.
Substituting the given values $n_P = n$ and $n_Q = \frac{n}{4}$:
$n^2 A_P^2 = (\frac{n}{4})^2 A_Q^2$.
$n^2 A_P^2 = \frac{n^2}{16} A_Q^2$.
$A_P^2 = \frac{A_Q^2}{16}$.
$A_Q^2 = 16 A_P^2$.
Taking the square root on both sides,we get:
$A_Q = 4 A_P$.

(A) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate.
$1$. The output of the $NOR$ gate is $\overline{A+B}$.
$2$. This output is then passed through a $NOT$ gate (inverter),which performs the complement operation.
$3$. The final output $Y$ is the complement of the $NOR$ output: $Y = \overline{(\overline{A+B})} = A+B$.
$4$. The Boolean expression $A+B$ corresponds to the $OR$ gate.
Therefore,the combination of a $NOR$ gate and a $NOT$ gate acts as an $OR$ gate.

(B) Let the initial lengths be $L_A$ and $L_B$ at temperature $T_0$. At a temperature $T = T_0 + \Delta T$,the new lengths are:
$L_A' = L_A(1 + \alpha_A \Delta T)$
$L_B' = L_B(1 + \alpha_B \Delta T)$
The difference in lengths is $\Delta L = L_A' - L_B' = (L_A - L_B) + (L_A \alpha_A - L_B \alpha_B) \Delta T$.
For the difference in length to remain constant at all temperatures,the term containing $\Delta T$ must be zero.
Therefore,$L_A \alpha_A - L_B \alpha_B = 0$.
$L_A \alpha_A = L_B \alpha_B$.
Thus,the ratio $\frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A}$.

(A) Carnot cycle consists of four reversible processes:
$1$. Isothermal expansion: The gas expands at a constant temperature $T_1$ by absorbing heat from the source.
$2$. Adiabatic expansion: The gas expands further without any heat exchange,and its temperature drops to $T_2$.
$3$. Isothermal compression: The gas is compressed at a constant temperature $T_2$ by rejecting heat to the sink.
$4$. Adiabatic compression: The gas is compressed further without any heat exchange,and its temperature rises back to $T_1$.
As shown in the $P-V$ diagram,the cycle starts with isothermal expansion from state $1$ to state $2$. Therefore,the correct option is $A$.

(C) When an observer moves towards a stationary source of sound,the apparent frequency $n^{\prime}$ is given by the Doppler effect formula:
$n^{\prime} = \left( \frac{v + v_{o}}{v} \right) n$
Where $v$ is the velocity of sound and $v_{o}$ is the velocity of the observer.
Given that $v_{o} = \frac{v}{5}$,we substitute this into the formula:
$n^{\prime} = \left( \frac{v + \frac{v}{5}}{v} \right) n = \left( \frac{\frac{6v}{5}}{v} \right) n = \frac{6}{5} n = 1.2n$
The increase in frequency is $\Delta n = n^{\prime} - n = 1.2n - n = 0.2n$.
The percentage increase is $\left( \frac{\Delta n}{n} \right) \times 100 = \left( \frac{0.2n}{n} \right) \times 100 = 20 \%$.

(C) Given: Velocity of sound $v = 330 \ m/s$,path difference $x = 40 \ cm = 0.4 \ m$,and phase difference $\phi = 1.6 \pi$.
We know the relationship between phase difference and path difference is given by $\phi = \frac{2 \pi}{\lambda} \cdot x$.
Substituting the values: $1.6 \pi = \frac{2 \pi}{\lambda} \times 0.4$.
$1.6 = \frac{0.8}{\lambda} \implies \lambda = \frac{0.8}{1.6} = 0.5 \ m$.
Using the wave equation $v = f \lambda$,we find the frequency $f = \frac{v}{\lambda}$.
$f = \frac{330}{0.5} = 660 \ Hz$.

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the wire.
When the bob is in air,the tension $T_1 = W = mg$,where $W$ is the weight of the bob.
So,$n_1 = \frac{1}{2L} \sqrt{\frac{W}{\mu}}$.
When the bob is immersed in water,the effective weight (tension) becomes $T_2 = W - F_B$,where $F_B$ is the buoyant force.
$T_2 = W - \frac{W}{\sigma} = W(1 - \frac{1}{\sigma})$,where $\sigma$ is the relative density of the metal.
So,$n_2 = \frac{1}{2L} \sqrt{\frac{W(1 - 1/\sigma)}{\mu}}$.
Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{1}{1 - 1/\sigma}} = \sqrt{\frac{\sigma}{\sigma - 1}}$.
Squaring both sides: $\frac{n_1^2}{n_2^2} = \frac{\sigma}{\sigma - 1}$.
$n_1^2(\sigma - 1) = n_2^2 \sigma$.
$\sigma(n_1^2 - n_2^2) = n_1^2$.
Therefore,$\sigma = \frac{n_1^2}{n_1^2 - n_2^2}$.

(D) The end correction $e$ for a tube of circular cross-section is given by the formula $e = 0.6r$ or $e = 0.3d$,where $r$ is the radius and $d$ is the diameter of the tube.
Since the end correction $e$ is directly proportional to the diameter $d$ $(e \propto d)$,increasing the diameter of the tube will result in a larger end correction.
Therefore,the end correction will be more if the tube is widened.

$(C)$ For an $fcc$ unit cell, the relationship between edge length $a$ and radius $r$ is $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.
Given $r = 106.05 \ pm$, we calculate $a = 2 \times 1.414 \times 106.05 \approx 300 \ pm$.
Converting to centimeters: $a = 300 \times 10^{-10} \ cm = 3 \times 10^{-8} \ cm$.
Volume of the unit cell $V = a^3 = (3 \times 10^{-8} \ cm)^3 = 27 \times 10^{-24} \ cm^3 = 2.7 \times 10^{-23} \ cm^3$.

(A) In a face-centered cubic $(fcc)$ lattice,each atom is in contact with $4$ atoms in its own layer,$4$ atoms in the layer above,and $4$ atoms in the layer below.
Therefore,the total coordination number of an atom in an $fcc$ structure is $4 + 4 + 4 = 12$.

(B) For a $bcc$ unit cell,the number of atoms per unit cell,$n = 2$.
The formula for density is $\rho = \frac{M \times n}{a^3 \times N_A}$.
Substituting the given values: $5.6 \ g \ cm^{-3} = \frac{M \times 2}{75 \ cm^3 \ mol^{-1}}$.
Solving for $M$: $M = \frac{5.6 \times 75}{2} = 210 \ g \ mol^{-1}$.

(B) The formula for density $(\rho)$ of a cubic unit cell is given by $\rho = \frac{M \times n}{a^3 \times N_A}$.
Here,$M$ is the molar mass,$n$ is the number of atoms per unit cell,$a^3$ is the volume of the unit cell,and $N_A$ is Avogadro's number.
Rearranging the formula to solve for $n$: $n = \frac{\rho \times a^3 \times N_A}{M}$.
Substituting the given values: $n = \frac{8.6 \ g \ cm^{-3} \times 21.5 \ cm^3 \ mol^{-1}}{92.0 \ g \ mol^{-1}}$.
$n = \frac{184.9}{92.0} \approx 2$.
Therefore,the number of atoms per unit cell is $2$.

(C) For a $bcc$ unit cell,the number of atoms per unit cell is $n = 2$.
The formula for density is $\rho = \frac{M \times n}{a^3 \times N_A}$.
Rearranging for the volume of the unit cell $(a^3)$,we get $a^3 = \frac{M \times n}{\rho \times N_A}$.
Substituting the given values: $a^3 = \frac{92 \ g \ mol^{-1} \times 2}{5.0 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}}$.
$a^3 = \frac{184}{5.0 \times 10^{24}} \ cm^3 = 36.8 \times 10^{-24} \ cm^3 = 3.68 \times 10^{-23} \ cm^3$.

(B) tetrahedral void is a type of interstitial space formed in a close-packed arrangement of spheres.
To create a tetrahedral void,a minimum of $4$ spheres are required.
The void is formed in the empty space between these $4$ spheres when they are arranged in a tetrahedral geometry.
In structures like $FCC$ (Face-Centered Cubic) or $HCP$ (Hexagonal Close-Packed),this void is created by the arrangement of $4$ spheres.

(D) For an $fcc$ structure,the number of atoms per unit cell,$n = 4$.
The formula for density is $\rho = \frac{n \times M}{a^3 \times N_{A}}$.
Given: $M = 197 \ g \ mol^{-1}$ and $a^3 \times N_{A} = 40 \ cm^3 \ mol^{-1}$.
Substituting the values: $\rho = \frac{4 \times 197 \ g \ mol^{-1}}{40 \ cm^3 \ mol^{-1}} = 19.7 \ g \ cm^{-3}$.

(B) For a simple cubic unit cell,the relationship between edge length $a$ and radius $r$ is $r = \frac{a}{2}$.
Therefore,$a = 2r = 2 \times 400 \ pm = 800 \ pm$.
Converting the edge length to centimeters: $a = 800 \ pm = 800 \times 10^{-10} \ cm = 8 \times 10^{-8} \ cm$.
The volume of the unit cell is $V = a^3 = (8 \times 10^{-8} \ cm)^3$.
$V = 512 \times 10^{-24} \ cm^3 = 5.12 \times 10^{-22} \ cm^3$.

(A) In a crystal lattice,the number of octahedral voids is equal to the number of atoms present in the lattice.
Given,the amount of compound $= 0.2 \ mol$.
Number of atoms $= 0.2 \times N_A = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$.
Since the number of octahedral voids equals the number of atoms,the number of octahedral voids $= 1.2044 \times 10^{23}$.

(A) For an $fcc$ unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by $r = \frac{a}{2\sqrt{2}}$.
Rearranging for $a$, we get $a = 2\sqrt{2} \times r$.
Given $r = 144 \ pm$, then $a = 2 \times 1.414 \times 144 \ pm = 407.23 \ pm$.
Converting to centimeters: $407.23 \ pm = 407.23 \times 10^{-10} \ cm = 4.07 \times 10^{-8} \ cm$.

(D) For $bcc$ unit cell,the number of atoms per unit cell $n = 2$.
The density formula is $\rho = \frac{M \times n}{a^3 \times N_A}$.
Given: $\rho = 10 \ g \ cm^{-3}$,$a = 4 \times 10^{-8} \ cm$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $10 = \frac{M \times 2}{(4 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$.
$M = \frac{10 \times 64 \times 10^{-24} \times 6.022 \times 10^{23}}{2}$.
$M = \frac{10 \times 64 \times 0.1 \times 6.022}{2} = \frac{385.4}{2} \approx 192.7 \ g \ mol^{-1}$.
Thus,the molar mass is approximately $193 \ g \ mol^{-1}$.

(A) For a $BCC$ unit cell, the relationship between the atomic radius $(r)$ and the edge length $(a)$ is given by: $4r = \sqrt{3}a$.
Substituting the given value of $a = 287 \ pm$:
$r = \frac{\sqrt{3}}{4} \times 287$
$r = \frac{1.732 \times 287}{4}$
$r = \frac{497.084}{4} \approx 124.27 \ pm$.
Rounding to the nearest integer, the atomic radius is $124 \ pm$.

(A) The seven crystal systems are Cubic,Orthorhombic,Tetragonal,Monoclinic,Triclinic,Rhombohedral,and Hexagonal.
By observing the Bravais lattices for these systems,we find that the 'Simple' (or primitive) unit cell is present in all seven crystal systems.
| Sr. No. | Type of System | Bravais lattices present |
| :--- | :--- | :--- |
| $1$ | Cubic | Simple,Face-centered,Body-centered |
| $2$ | Orthorhombic | Simple,Base-centered,Face-centered,Body-centered |
| $3$ | Tetragonal | Simple,Body-centered |
| $4$ | Monoclinic | Simple,Base-centered |
| $5$ | Triclinic | Simple |
| $6$ | Rhombohedral | Simple |
| $7$ | Hexagonal | Simple or primitive |

(A) For a simple cubic unit cell,the number of atoms per unit cell $(n)$ is $1$.
Total volume occupied by atoms in the unit cell $= n \times \frac{4}{3} \pi r^3$.
Given $r = 3 \times 10^{-8} \ cm$.
Volume $= 1 \times \frac{4}{3} \times 3.14159 \times (3 \times 10^{-8} \ cm)^3$.
Volume $= \frac{4}{3} \times 3.14159 \times 27 \times 10^{-24} \ cm^3$.
Volume $= 4 \times 3.14159 \times 9 \times 10^{-24} \ cm^3$.
Volume $= 113.097 \times 10^{-24} \ cm^3 = 1.13 \times 10^{-22} \ cm^3$.

(C) For $BCC$ unit cell,the relation between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3}}{4} a$.
The volume of one spherical particle is given by $V = \frac{4}{3} \pi r^3$.
Substituting the value of $r$:
$V = \frac{4}{3} \pi \left( \frac{\sqrt{3}}{4} a \right)^3 = \frac{4}{3} \pi \left( \frac{3 \sqrt{3} a^3}{64} \right) = \frac{\sqrt{3} \pi a^3}{16}$.

(B) The density formula is given by $\rho = \frac{n \times M}{a^3 \times N_A}$.
Rearranging for the number of atoms per unit cell $(n)$: $n = \frac{\rho \times N_A \times a^3}{M}$.
Given: $\rho = 4 \ g \ cm^{-3}$,$M = 149 \ g \ mol^{-1}$,$a = 5 \mathring{A} = 5 \times 10^{-8} \ cm$,and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values: $n = \frac{4 \times 6.022 \times 10^{23} \times (5 \times 10^{-8})^3}{149}$.
$n = \frac{4 \times 6.022 \times 10^{23} \times 125 \times 10^{-24}}{149} = \frac{301.1}{149} \approx 2.02$.
Since $n \approx 2$,the crystal structure is Body-centred cubic $(BCC)$.

(D) In a $ccp$ structure,the number of effective atoms per unit cell is $4$. Since $B^{-}$ ions form the $ccp$ structure,the number of $B^{-}$ ions $= 4$.
In a $ccp$ lattice,the number of tetrahedral voids is twice the number of atoms,so the number of tetrahedral voids $= 2 \times 4 = 8$.
$A^{+}$ ions occupy half of the tetrahedral voids,so the number of $A^{+}$ ions $= 8 \times (1/2) = 4$.
The ratio of $A^{+} : B^{-}$ is $4 : 4$,which simplifies to $1 : 1$.
Therefore,the formula of the solid is $AB$.

(A) For an $fcc$ unit cell,the packing efficiency is $74 \%$.
Therefore,the percentage of unoccupied space (void volume) is $100 \% - 74 \% = 26 \%$.
Void volume $= 1.25 \times 10^{-22} \ cm^3 \times \frac{26}{100}$.
Void volume $= 3.25 \times 10^{-23} \ cm^3$.

(C) The mass of the unit cell is given by the product of density $(\rho)$ and volume $(a^3)$,which is $\rho \times a^3 = 3 \times 10^{-22} \ g$.
Number of unit cells = $\frac{\text{Total mass of metal}}{\text{Mass of one unit cell}}$
Number of unit cells = $\frac{0.9 \ g}{3 \times 10^{-22} \ g}$
Number of unit cells = $0.3 \times 10^{22} = 3.0 \times 10^{21}$.

(C) In a base-centred unit cell,particles are present at the $8$ corners and at the centers of $2$ opposite faces.
Contribution from $8$ corners $= 8 \times \frac{1}{8} = 1$.
Contribution from $2$ face centres $= 2 \times \frac{1}{2} = 1$.
Total number of particles $= 1 + 1 = 2$.

(C) The mass of the metal is given as $m = 0.4 \ g$.
The product of density $(\rho)$ and volume of the unit cell $(V = a^3)$ is given as $\rho \times a^3 = 1.2 \times 10^{-22} \ g$.
The number of unit cells is calculated by dividing the total mass of the metal by the mass of one unit cell.
$\text{Number of unit cells} = \frac{\text{Total mass}}{\text{Mass of one unit cell}} = \frac{m}{\rho \times a^3}$.
Substituting the values: $\frac{0.4 \ g}{1.2 \times 10^{-22} \ g} = 3.33 \times 10^{21}$.
Therefore,the number of unit cells is $3.3 \times 10^{21}$.

(B) In a crystal lattice,the number of tetrahedral voids is twice the number of atoms present in the lattice.
$1 \ mole$ of compound contains $6.022 \times 10^{23}$ atoms.
$0.6 \ mole$ of compound contains $0.6 \times 6.022 \times 10^{23}$ atoms.
Number of tetrahedral voids $= 2 \times (\text{Number of atoms}) = 2 \times 0.6 \times 6.022 \times 10^{23}$.
$= 1.2 \times 6.022 \times 10^{23} = 7.2264 \times 10^{23}$ voids.

(C) The correct answer is $C$ (Schottky defect).
$A$ $Schottky$ defect occurs in an ionic solid when equal numbers of cations and anions are missing from their regular positions in the crystal lattice,creating vacancies.
This defect maintains the overall electrical neutrality of the crystal.

(C) The magnetic properties of the given substances are as follows:
$1$. $NaCl$: Diamagnetic
$2$. $C_6H_6$: Diamagnetic
$3$. $CrO_2$: Ferromagnetic
$4$. $H_2O$: Diamagnetic
Therefore,$CrO_2$ is the ferromagnetic substance.

(A) An $n$-type semiconductor is formed when a group $14$ element (like silicon or germanium) is doped with a group $15$ element (like phosphorus or arsenic) that has more valence electrons.
$(1)$ Silicon has $4$ valence electrons. Phosphorus has $5$ valence electrons.
$(2)$ When silicon is doped with phosphorus,$4$ electrons of phosphorus form covalent bonds with silicon,while the $5^{th}$ electron remains free.
$(3)$ This extra free electron increases the electrical conductivity,resulting in an $n$-type semiconductor.

(D) Ferromagnetic substances are those that are strongly attracted by a magnetic field and show permanent magnetism even when the magnetic field is removed.
Examples include $Fe$,$Co$,$Ni$,$CrO_2$,and $Fe_3O_4$.
Analyzing the given options:
- $NaCl$ is diamagnetic.
- $H_2O$ is diamagnetic.
- $O_2$ is paramagnetic.
- $CrO_2$ is ferromagnetic.
Therefore,the correct option is $D$.

(B) solution of a solid solute in a liquid solvent is a common type of solution.
In the given options,$A$ (Sea water) is a solution of salt (solid) in water (liquid).
$B$ (Sugar in water) is also a solid in liquid solution.
$C$ (Carbonated water) is a gas in liquid solution.
$D$ (Chloroform in nitrogen) is a liquid in gas solution.
Since the question asks for a solid in liquid solution,both $A$ and $B$ fit the criteria. However,$B$ is the most standard textbook example.

(A) The solubility of a polar solute in a polar solvent is mainly due to the favorable interactions between the solute and solvent molecules. These interactions are crucial for the dissolution process.
The most appropriate explanation for this is when the solute - solute,solvent - solvent,and solute - solvent interactions are comparable in magnitude. This balance allows the solute to dissolve efficiently,as the solvent can break apart the solute molecules while maintaining similar intermolecular forces within the mixture.
Correct answer: $A$. Solute - solute interactions,solute - solvent interactions,and solvent - solvent interactions are of similar magnitude.

(D) An alloy is a homogeneous mixture of two or more metals,or a metal and a non-metal. Since both the solute and the solvent are in the solid state,an alloy is classified as a $Solid$ in $Solid$ solution.

(D) Among the given options,sodium sulfate $(Na_2SO_4)$ exhibits a decrease in solubility with an increase in temperature.
This behavior is unusual because,for most ionic solids,solubility tends to increase with temperature.
However,for $Na_2SO_4$,its dissolution is exothermic,meaning it releases heat when it dissolves in water.
According to Le Chatelier's principle,if the dissolution process is exothermic,increasing the temperature will shift the equilibrium to favor the undissolved solid,thereby reducing solubility at higher temperatures.
Solubility trends for other salts:
- $(1)$ $NaCl$ (Sodium chloride): $NaCl$'s solubility increases with temperature.
- $(2)$ $KNO_3$ (Potassium nitrate): $KNO_3$'s solubility increases significantly with temperature.
- $(3)$ $NaNO_3$ (Sodium nitrate): $NaNO_3$'s solubility also increases with temperature.

(D) Since both solutions boil at the same temperature,their elevation in boiling point $(\Delta T_b)$ must be the same.
Since $\Delta T_b = K_b \times m$,and $K_b$ is the same for the same solvent (water),the molality $(m)$ of both solutions must be equal.
Given concentration in $g \ dm^{-3}$ is $C = \frac{W}{V}$. For a fixed volume of solvent,the molarity and molality are proportional to the concentration divided by the molar mass.
$\frac{C_{\text{glucose}}}{M_{\text{glucose}}} = \frac{C_A}{M_A}$
Substituting the given values:
$\frac{18}{180} = \frac{6}{M_A}$
$0.1 = \frac{6}{M_A}$
$M_A = \frac{6}{0.1} = 60 \ g \ mol^{-1}$.

(C) Two solutions separated by a semipermeable membrane will not show any net flow of solvent if they are isotonic,meaning they have the same osmotic pressure $(\pi = CRT)$.
For solutions with the same volume,this condition is met if the number of moles of solute is equal $(n_{urea} = n_{sucrose})$.
Checking option $C$:
Moles of urea = $\frac{6 \ g}{60 \ g \ mol^{-1}} = 0.1 \ mol$.
Moles of sucrose = $\frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$.
Since the number of moles is equal,the osmotic pressures are equal,and no net flow of solvent occurs.

(A) Carbonated water (soda-water) is prepared by dissolving $CO_2$ gas under high pressure in water.
Since the solute is a gas $(CO_2)$ and the solvent is a liquid (water), it is classified as a $Gas \ in \ liquid$ type of solution.

(B) solution of iodine in air is an example of a solid dispersed in a gas. Iodine is a solid at room temperature and it undergoes sublimation to form iodine vapor,which is a gas. Therefore,it is a solid in gas type of solution.

(C) According to Raoult's Law for non-volatile solutes: $\frac{P^0 - P}{P^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W_2 \times M_1}{M_2 \times W_1}$
Given: $W_2 = 2 \ g$,$W_1 = 50 \ g$,$M_2 = 64 \ g \ mol^{-1}$,$M_1 = 78 \ g \ mol^{-1}$,$P^0 = 640 \ mm \ Hg$
$\frac{640 - P}{640} = \frac{2 \times 78}{64 \times 50}$
$\frac{640 - P}{640} = \frac{156}{3200} = 0.04875$
$640 - P = 640 \times 0.04875 = 31.2$
$P = 640 - 31.2 = 608.8 \ mm \ Hg$
The closest value is $608.64 \ mm \ Hg$.

(C) The mole fraction of $A$ is $x_A = \frac{n_A}{n_A + n_B} = \frac{2}{2+3} = 0.4$.
The mole fraction of $B$ is $x_B = \frac{n_B}{n_A + n_B} = \frac{3}{2+3} = 0.6$.
According to Raoult's law,the total vapour pressure of the solution is $P_{\text{total}} = P_{A}^{\circ} x_A + P_{B}^{\circ} x_B$.
Substituting the values: $P_{\text{total}} = (420 \times 0.4) + (610 \times 0.6)$.
$P_{\text{total}} = 168 + 366 = 534 \ mm \ Hg$.

(A) The relation between the vapour pressure of the solution $(P_1)$,the vapour pressure of the pure solvent $(P_1^*)$,and the mole fraction of the solvent $(x_1)$ in the solution is defined by Raoult's Law for an ideal solution.
Raoult's Law states that for a solution of volatile liquids,the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
The mathematical expression is:
$P_1 = P_1^* \cdot x_1$
Where:
-$P_1$ is the partial vapour pressure of the solvent in the solution.
-$P_1^*$ is the vapour pressure of the pure solvent.
-$x_1$ is the mole fraction of the solvent in the solution.
Therefore,the correct option is $(A)$.

(B) The formula for osmotic pressure is $\pi = MRT$.
Here,$M$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.
Molarity $M = \frac{n_2}{V} = \frac{\text{mass} / \text{molar mass}}{V} = \frac{3 / 60}{2} = \frac{0.05}{2} = 0.025 \ mol \ dm^{-3}$.
Substituting the values: $\pi = 0.025 \times 0.0821 \times 300$.
$\pi = 0.61575 \ atm \approx 0.62 \ atm$.

(D) $(1)$ Ideal solutions obey Raoult's law over the entire range of temperature and concentration: This is true. An ideal solution is one in which the intermolecular interactions between the components are identical to those between the molecules of the individual components.
$(2)$ For an ideal solution,$\Delta_{\text{mix}}V = 0$: This is true. In an ideal solution,the volume change upon mixing is zero,meaning the total volume of the solution is the sum of the volumes of the individual components.
$(3)$ Non-ideal solutions do not obey Raoult's law over the entire range of concentration: This is true. Non-ideal solutions exhibit deviations from Raoult's law due to differences in intermolecular forces between the components.
$(4)$ The vapour pressure of a non-ideal solution always lies between the vapour pressures of the pure components: This statement is false. In non-ideal solutions,the vapour pressure can be higher (positive deviation) or lower (negative deviation) than the vapour pressures of the pure components.

(C) The formula for osmotic pressure is $\pi = CRT = \frac{W_2}{M_2 V} RT$.
Rearranging to solve for molar mass $M_2$:
$M_2 = \frac{W_2 RT}{\pi V}$
Given: $W_2 = 0.8 \ g$,$V = 0.3 \ dm^3$,$\pi = 0.2 \ atm$,$T = 300 \ K$,$R = 0.082 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$.
Substituting the values:
$M_2 = \frac{0.8 \times 0.082 \times 300}{0.2 \times 0.3}$
$M_2 = \frac{19.68}{0.06} = 328 \ g \ mol^{-1}$.

(A) According to Henry's Law,the solubility $(S)$ of a gas is given by the formula: $S = K_{H} \times P$.
Given: $K_{H} = 0.16 \ mol \ dm^{-3} \ bar^{-1}$ and $P = 0.15 \ bar$.
Substituting the values: $S = 0.16 \ mol \ dm^{-3} \ bar^{-1} \times 0.15 \ bar = 0.024 \ mol \ dm^{-3}$.
Therefore,$S = 2.4 \times 10^{-2} \ mol \ dm^{-3}$.

(A) According to Henry's law,the solubility $(S)$ is related to the partial pressure $(P)$ by the equation: $S = K_H P$
Here,$K_H$ is the Henry's law constant.
Rearranging the formula to solve for $K_H$: $K_H = \frac{S}{P}$
Substituting the given values: $K_H = \frac{5.14 \times 10^{-4} \ mol \ dm^{-3}}{0.75 \ bar} = 6.85 \times 10^{-4} \ mol \ dm^{-3} \ bar^{-1}$

(C) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \times m$.
Given: $\Delta T_{f} = 0.93 \ K$ and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Rearranging the formula for molality $(m)$:
$m = \frac{\Delta T_{f}}{K_{f}} = \frac{0.93 \ K}{1.86 \ K \ kg \ mol^{-1}} = 0.5 \ mol \ kg^{-1}$.

(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 119.6^{\circ} C - 118^{\circ} C = 1.6 \ K$.
Using the formula $\Delta T_{b} = \frac{1000 \times K_{b} \times W_{2}}{M_{2} \times W_{1}}$,where $W_{2} = 5 \ g$,$W_{1} = 50 \ g$,and $K_{b} = 3.2 \ K \ kg \ mol^{-1}$.
Rearranging for molar mass $M_{2}$: $M_{2} = \frac{1000 \times 3.2 \times 5}{1.6 \times 50}$.
$M_{2} = \frac{16000}{80} = 200 \ g \ mol^{-1}$.

(D) The formula for freezing point depression is $\Delta T_{f} = i \times m \times K_{f}$.
Given $\Delta T_{f} = 0.7 \ K$,$m = 0.2 \ m$,and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.7 = i \times 0.2 \times 1.86$.
$i = \frac{0.7}{0.2 \times 1.86} = \frac{0.7}{0.372} \approx 1.882$.

(D) The molal elevation constant $(K_b)$ is defined as the elevation in boiling point produced by a $1$ molal solution (i.e.,$1 \ mol$ of solute dissolved in $1 \ kg$ of solvent). Option $D$ states '$1$ molar solution',which is incorrect because molarity depends on temperature,whereas molality is used for boiling point elevation.

(B) The formula for molar mass of a solute is: $M_2 = \frac{1000 \times K_{f} \times W_2}{\Delta T_{f} \times W_1}$
Given: $W_2 = 1.5 \ g$,$W_1 = 90 \ g$,$\Delta T_{f} = 0.25 \ K$,$K_{f} = 1.2 \ K \ kg \ mol^{-1}$.
Substituting the values: $M_2 = \frac{1000 \times 1.2 \times 1.5}{0.25 \times 90}$
$M_2 = \frac{1800}{22.5} = 80 \ g \ mol^{-1}$.

(B) Calculate the molarity $(M)$ of the urea solution: $M_{urea} = \frac{6 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = 0.1 \ mol \ L^{-1}$.
Calculate the molarity $(M)$ of the sucrose solution: $M_{sucrose} = \frac{17.12 \ g \ L^{-1}}{342 \ g \ mol^{-1}} \approx 0.05 \ mol \ L^{-1}$.
Since osmotic pressure $\pi = CRT$,and the temperature $(T)$ is constant,the solution with higher molar concentration has higher osmotic pressure.
As $0.1 \ M > 0.05 \ M$,the urea solution has a higher osmotic pressure than the sucrose solution.
Therefore,the urea solution is hypertonic to the sucrose solution.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
Given $\Delta T_{b} = 1.75 \ K$ and $K_{b} = 3.5 \ K \ kg \ mol^{-1}$.
Substituting the values: $1.75 \ K = 3.5 \ K \ kg \ mol^{-1} \times m$.
Solving for molality: $m = \frac{1.75}{3.5} = 0.50 \ mol \ kg^{-1}$.

(A) The elevation in boiling point $(\Delta T_b)$ is given by the formula: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent $(W_1 \text{ in grams})$.
$m = \frac{n_2 \times 1000}{W_1} = \frac{W_2 \times 1000}{M_2 \times W_1}$.
Substituting this into the elevation formula: $\Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1}$.
Rearranging for the molar mass of the solute $(M_2)$: $M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}$.