AP EAMCET 2025 Chemistry Question Paper with Answer and Solution

(C) The given circuit consists of a $NOR$ gate, a $NAND$ gate with its inputs tied together (acting as a $NOT$ gate), and a $NOT$ gate.
Let the inputs be $A$ and $B$.
The output of the $NOR$ gate is $Y_1 = \overline{A+B}$.
The $NAND$ gate with tied inputs acts as a $NOT$ gate, so its output is $Y_2 = \overline{Y_1 \cdot Y_1} = \overline{Y_1}$.
Substituting $Y_1$, we get $Y_2 = \overline{\overline{A+B}} = A+B$.
The final output $y$ passes through a $NOT$ gate, so $y = \overline{Y_2} = \overline{A+B}$.
This is the Boolean expression for a $NOR$ gate.
Therefore, the circuit is equivalent to a $NOR$ gate.

(C) Let the position vectors be $\vec{a} = \hat{i}+2\hat{j}+2\hat{k}$,$\vec{b} = 2\hat{i}-\hat{j}$,$\vec{c} = \hat{i}+\hat{j}+3\hat{k}$,and $\vec{d} = 4\hat{j}+5\hat{k}$.
First,find the vectors representing the sides:
$\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (-1-2)\hat{j} + (0-2)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$.
$\vec{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (1-(-1))\hat{j} + (3-0)\hat{k} = -\hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{CD} = \vec{d} - \vec{c} = (0-1)\hat{i} + (4-1)\hat{j} + (5-3)\hat{k} = -\hat{i} + 3\hat{j} + 2\hat{k}$.
$\vec{DA} = \vec{a} - \vec{d} = (1-0)\hat{i} + (2-4)\hat{j} + (2-5)\hat{k} = \hat{i} - 2\hat{j} - 3\hat{k}$.
Note that $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$,which confirms $ABCD$ is a parallelogram.
Now check the lengths of the sides:
$|\vec{AB}| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{1+9+4} = \sqrt{14}$.
$|\vec{BC}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$.
Since adjacent sides are equal $(|\vec{AB}| = |\vec{BC}| = \sqrt{14})$,the parallelogram is a rhombus.

(A) Let the origin be $O(0,0,0)$ and the foot of the perpendicular be $P(2,-1,3)$.
Since $OP$ is the normal to the plane,the direction ratios of the normal are the same as the direction ratios of the line $OP$.
The direction ratios of $OP$ are $(2-0, -1-0, 3-0) = (2, -1, 3)$.
Thus,the normal vector to the plane is $\vec{n} = 2\hat{i} - \hat{j} + 3\hat{k}$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Substituting the point $P(2, -1, 3)$ and the normal vector $(2, -1, 3)$ into the equation:
$2(x-2) - 1(y-(-1)) + 3(z-3) = 0$
$2(x-2) - 1(y+1) + 3(z-3) = 0$
$2x - 4 - y - 1 + 3z - 9 = 0$
$2x - y + 3z - 14 = 0$
Therefore,the equation of the plane is $2x - y + 3z - 14 = 0$.

(C) Let the mass of $CaCO_3$ be $x \ g$ and the mass of $MgCO_3$ be $(1.84 - x) \ g$.
Heating the carbonates results in the following reactions:
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
$MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$
The molar mass of $CaCO_3 = 100 \ g/mol$,$CaO = 56 \ g/mol$,$MgCO_3 = 84 \ g/mol$,and $MgO = 40 \ g/mol$.
The mass of the residue $(CaO + MgO)$ is $0.96 \ g$.
Mass of $CaO = (x / 100) \times 56 = 0.56x$.
Mass of $MgO = ((1.84 - x) / 84) \times 40 = 0.476(1.84 - x)$.
$0.56x + 0.476(1.84 - x) = 0.96$.
$0.56x + 0.876 - 0.476x = 0.96$.
$0.084x = 0.084$.
$x = 1 \ g$.
Percentage of $CaCO_3 = (1 / 1.84) \times 100 \approx 54.34\%$.

(B) Let the element be $M$ and its valency be $n$. The chloride formed is $MCl_n$.
Mass of $Cl$ in $315.5 \ g$ of $MCl_n = 315.5 \ g - 209 \ g = 106.5 \ g$.
Equivalent mass of $Cl = \frac{35.5}{1} = 35.5 \ g$.
Number of equivalents of $Cl = \frac{106.5}{35.5} = 3$.
Since equivalents of $M$ = equivalents of $Cl$,equivalents of $M = 3$.
Equivalent mass of $M = \frac{\text{Mass of } M}{\text{Equivalents}} = \frac{209}{3} \ g$.
Now,for the oxide $M_xO_y$,equivalents of $M$ = equivalents of $O$.
For $418 \ g$ of $M$,equivalents of $M = \frac{418}{209/3} = 418 \times \frac{3}{209} = 2 \times 3 = 6$.
Equivalent mass of $O = \frac{16}{2} = 8 \ g$.
Mass of $O = \text{Equivalents} \times \text{Equivalent mass} = 6 \times 8 = 48 \ g$.

(D) $1$. Calculate total moles of the mixture $(n_{total})$ using the ideal gas law: $PV = nRT$. Given $P = 1 \ atm$,$V = 28 \ L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 273 \ K$. $n_{total} = \frac{1 \times 28}{0.0821 \times 273} \approx 1.25 \ mol$.
$2$. Let $x$ be the moles of $C_2H_6$ and $y$ be the moles of $C_2H_4$. So,$x + y = 1.25$.
$3$. Combustion reactions:
$C_2H_6 + 3.5O_2 \rightarrow 2CO_2 + 3H_2O$
$C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$
$4$. Moles of $O_2$ required: $3.5x + 3y = \frac{128}{32} = 4$.
$5$. Solve the system of equations: $x + y = 1.25 \Rightarrow x = 1.25 - y$.
$6$. Substitute into the second equation: $3.5(1.25 - y) + 3y = 4$ $\Rightarrow 4.375 - 3.5y + 3y = 4$ $\Rightarrow 0.5y = 0.375$ $\Rightarrow y = 0.75$.
$7$. Mole fraction of $C_2H_4 = \frac{y}{x+y} = \frac{0.75}{1.25} = 0.6$.

(A) The balanced chemical equation for the combustion of ethane $(C_2H_6)$ is:
$2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$
Calculate the moles of reactants:
Molar mass of $C_2H_6 = 30 \ g/mol$. Moles of $C_2H_6 = 15 \ g / 30 \ g/mol = 0.5 \ mol$.
Molar mass of $O_2 = 32 \ g/mol$. Moles of $O_2 = 112 \ g / 32 \ g/mol = 3.5 \ mol$.
Determine the limiting reagent:
For $0.5 \ mol$ of $C_2H_6$,required $O_2 = (7/2) \times 0.5 = 1.75 \ mol$.
Since we have $3.5 \ mol$ of $O_2$,$C_2H_6$ is the limiting reagent.
Calculate moles of products and remaining reactants:
$C_2H_6$ consumed = $0.5 \ mol$ (Remaining = $0 \ mol$).
$O_2$ consumed = $1.75 \ mol$ (Remaining = $3.5 - 1.75 = 1.75 \ mol$).
$CO_2$ produced = $(4/2) \times 0.5 = 1.0 \ mol$.
$H_2O$ produced = $(6/2) \times 0.5 = 1.5 \ mol$.
Total moles of gaseous substances = $n(O_2) + n(CO_2) + n(H_2O) = 1.75 + 1.0 + 1.5 = 4.25 \ mol$.

(C) Let the mass of $NaCl$ be $x \ g$ and the mass of $NaBr$ be $(4.0 - x) \ g$.
The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.
The molar mass of $NaBr = 23 + 80 = 103 \ g/mol$.
The mass of $Na$ in $NaCl = x \times (23 / 58.5) \ g$.
The mass of $Na$ in $NaBr = (4.0 - x) \times (23 / 103) \ g$.
Total mass of $Na = 30 \% \text{ of } 4.0 \ g = 1.2 \ g$.
So,$x(23 / 58.5) + (4.0 - x)(23 / 103) = 1.2$.
Dividing by $23$: $x / 58.5 + (4.0 - x) / 103 = 1.2 / 23 \approx 0.05217$.
$0.01709x + 0.03883 - 0.00971x = 0.05217$.
$0.00738x = 0.01334$.
$x \approx 1.807 \ g$.
Percentage of $NaCl = (1.807 / 4.0) \times 100 \approx 45.17 \%$.
The closest option is $45 \%$.

(D) For $I) 0.0063$: Leading zeros are not significant. The significant figures are $6$ and $3$. Total = $2$.
For $II) 132.00$: Trailing zeros in a number with a decimal point are significant. Total = $5$.
For $III) 1004$: Zeros between non-zero digits are significant. Total = $4$.
Therefore,the number of significant figures is $2, 5, 4$.

(D) For $A = 0.0025$: Leading zeros are not significant. The significant figures are $2$ and $5$. Total = $2$.
For $B = 500.0$: Trailing zeros in a number with a decimal point are significant. The significant figures are $5, 0, 0, 0$. Total = $4$.
For $C = 2.0034$: Zeros between non-zero digits are significant. The significant figures are $2, 0, 0, 3, 4$. Total = $5$.
Therefore,the number of significant figures are $2, 4, 5$.

(C) The ideal gas equation is given by $PV = nRT$.
For option $A$: If $n$ and $T$ are constant, then $PV = \text{constant}$, which means $P \propto 1/V$. The graph of $P$ versus $V$ should be a rectangular hyperbola, not a straight line.
For option $B$: If $V$ and $n$ are constant, then $P = (nR/V) \times T$. Since $nR/V$ is constant, $P \propto T$. The graph of $P$ versus $T$ should be a straight line passing through the origin, not a line with a negative slope.
For option $C$: If $V$ and $T$ are constant, then $P = (RT/V) \times n$. Since $RT/V$ is constant, $P \propto n$. The graph of $P$ versus $n$ is a straight line passing through the origin. This is correct.
For option $D$: If $P$ and $n$ are constant, then $V = (nR/P) \times T$. Thus $V \propto T$. The graph of $V$ versus $1/T$ would not be a straight line passing through the origin; rather, $V$ versus $T$ would be.
Therefore, the correct graph is $C$.

(C) For an ideal gas,the equation is $PV = nRT$.
For an isochore (constant volume process),the equation can be written as $P = (\frac{nR}{V})T$.
The slope of the $P$ vs $T$ graph is given by $m = \frac{nR}{V}$.
Given $n = 1 \ mol$ and slope $m = 0.082 \ atm \ K^{-1}$.
So,$\frac{1 \times 0.082}{V} = 0.082$,which implies $V = 1 \ L$.
Now,at $T = 12.2 \ K$,the pressure $P$ is calculated as $P = \frac{nRT}{V} = \frac{1 \times 0.082 \times 12.2}{1} = 1.0004 \ atm \approx 1.0 \ atm$.

(C) Statement-$I$ is correct: Thermal energy promotes random motion of particles,while intermolecular forces hold them together. If thermal energy dominates,the particles move apart,leading to a gaseous state.
Statement-$II$ is correct: From the ideal gas equation,$PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$. Rearranging gives $P = \frac{m}{V} \times \frac{RT}{M}$,where $\frac{m}{V} = d$ (density). Thus,$P = d \times \frac{RT}{M}$. At constant temperature $(T)$ and molar mass $(M)$,$P \propto d$ or $d \propto P$.

(B) Given: Total pressure $P_{total} = 2 \ bar$,Partial pressure of $H_2$ $(P_{H_2})$ = $1.778 \ bar$.
Partial pressure of $O_2$ $(P_{O_2})$ = $P_{total} - P_{H_2} = 2 - 1.778 = 0.222 \ bar$.
According to Dalton's Law,the mole fraction $(x)$ is proportional to the partial pressure: $x_{H_2} = \frac{P_{H_2}}{P_{total}} = \frac{1.778}{2} = 0.889$ and $x_{O_2} = \frac{P_{O_2}}{P_{total}} = \frac{0.222}{2} = 0.111$.
Assuming $1 \ mole$ of the mixture,moles of $H_2$ $(n_{H_2})$ = $0.889 \ mol$ and moles of $O_2$ $(n_{O_2})$ = $0.111 \ mol$.
Mass of $H_2$ = $n_{H_2} \times M_{H_2} = 0.889 \times 2 = 1.778 \ g$.
Mass of $O_2$ = $n_{O_2} \times M_{O_2} = 0.111 \times 32 = 3.552 \ g$.
Total mass = $1.778 + 3.552 = 5.33 \ g$.
Weight percentage of $H_2$ = $(\frac{1.778}{5.33}) \times 100 \approx 33.33 \%$.

(A) Let the mass of $H_2$ be $w \ g$ and the mass of $O_2$ be $2w \ g$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{w}{2}$.
Number of moles of $O_2$ $(n_{O_2})$ = $\frac{2w}{32} = \frac{w}{16}$.
Total moles $(n_{total})$ = $\frac{w}{2} + \frac{w}{16} = \frac{8w + w}{16} = \frac{9w}{16}$.
Mole fraction of $O_2$ $(x_{O_2})$ = $\frac{n_{O_2}}{n_{total}} = \frac{w/16}{9w/16} = \frac{1}{9}$.
Partial pressure of $O_2$ = $x_{O_2} \times P_{total} = \frac{1}{9} \times p = \frac{p}{9}$ atm.

(A) Using the ideal gas law $PV = nRT$,we find the moles of each gas at constant temperature $T = 300 \ K$.
For $H_2$: $n_1 = \frac{P_1 V_1}{RT} = \frac{1 \ bar \times 1 \ L}{RT} = \frac{1}{RT}$.
For $O_2$: $n_2 = \frac{P_2 V_2}{RT} = \frac{2 \ bar \times 2 \ L}{RT} = \frac{4}{RT}$.
Total moles $n_{total} = n_1 + n_2 = \frac{1+4}{RT} = \frac{5}{RT}$.
In the $10 \ L$ flask,the total pressure $P_{total}$ is given by $P_{total} = \frac{n_{total} RT}{V_{final}}$.
$P_{total} = \frac{(5/RT) \times RT}{10 \ L} = \frac{5}{10} \ bar = 0.5 \ bar$.

(D) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.
Since the time $t$ is the same,the rate of diffusion is directly proportional to the mass $w$ diffused: $r = \frac{w}{t}$.
Therefore,$\frac{w_1}{w_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $w_{H_2} = 2.0 \ g$,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$.
Substituting the values: $\frac{2.0}{w_{O_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
$w_{O_2} = 2.0 \times 4 = 8.0 \ g$.

(A) For an ideal gas,the product of pressure and volume $(PV)$ is proportional to the number of moles $(n)$ at a constant temperature $(T)$.
Using the principle of conservation of moles,the total pressure of the mixture $(P_{mix})$ in the final volume $(V_{final} = 100 \ L)$ is given by the sum of the partial pressures of $A$ and $B$.
Initial state of $A$: $P_1 = 1 \ bar$,$V_1 = 2 \ L$.
Initial state of $B$: $P_2 = p_B \ bar$,$V_2 = 4 \ L$.
Final state of mixture: $P_{mix} = 0.1 \ bar$,$V_{final} = 100 \ L$.
According to Boyle's Law $(P_1V_1 + P_2V_2 = P_{mix}V_{final})$:
$(1 \ bar \times 2 \ L) + (p_B \ bar \times 4 \ L) = 0.1 \ bar \times 100 \ L$
$2 + 4p_B = 10$
$4p_B = 10 - 2$
$4p_B = 8$
$p_B = 2 \ bar$.
Therefore,the value of $p_B$ is $2$.

(B) The relationship between $rms$ speed $(v_{rms})$ and most probable velocity $(v_{mp})$ is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$ and $v_{mp} = \sqrt{\frac{2RT}{M}}$.
Dividing the two expressions,we get: $\frac{v_{mp}}{v_{rms}} = \sqrt{\frac{2}{3}}$.
Given $v_{rms} = 3.16 \times 10^2 \ ms^{-1}$,we calculate $v_{mp} = v_{rms} \times \sqrt{\frac{2}{3}}$.
$v_{mp} = 3.16 \times 10^2 \times \sqrt{0.6667} \approx 3.16 \times 10^2 \times 0.8165$.
$v_{mp} \approx 2.58 \times 10^2 \ ms^{-1}$.

(B) The most probable speed is given by the formula $u_{mp} = \sqrt{\frac{2RT}{M}}$.
Given $u_{mp} = 200 \ ms^{-1}$ and molar mass $M = 2 \times 10^{-3} \ kg \ mol^{-1}$.
$200 = \sqrt{\frac{2RT}{2 \times 10^{-3}}} \implies 40000 = \frac{2RT}{2 \times 10^{-3}} \implies RT = 40 \ J \ mol^{-1}$.
The kinetic energy of an ideal gas is given by $KE = \frac{3}{2} nRT$.
Number of moles $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \ g}{2 \ g \ mol^{-1}} = 4 \ mol$.
$KE = \frac{3}{2} \times 4 \times 40 = 6 \times 40 = 240 \ J$.

(C) The formula for $rms$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
For $v_{rms}$ of $SO_2$ to be equal to $v_{rms}$ of $O_2$,we have $\sqrt{\frac{3RT_{SO_2}}{M_{SO_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.
Squaring both sides and canceling common terms,we get $\frac{T_{SO_2}}{M_{SO_2}} = \frac{T_{O_2}}{M_{O_2}}$.
Given $T_{O_2} = 27 + 273 = 300 \ K$,$M_{O_2} = 32 \ g/mol$,and $M_{SO_2} = 64 \ g/mol$.
Substituting the values: $\frac{T_{SO_2}}{64} = \frac{300}{32}$.
$T_{SO_2} = \frac{300 \times 64}{32} = 300 \times 2 = 600 \ K$.

(D) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} nRT$.
For $H_2$:
Molar mass $M(H_2) = 2 \ g/mol$.
Number of moles $n_1 = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$.
Temperature $T_1 = 27 + 273 = 300 \ K$.
$x = \frac{3}{2} \times 2 \times R \times 300 = 900R$.
For $O_2$:
Molar mass $M(O_2) = 32 \ g/mol$.
Number of moles $n_2 = \frac{6.4 \ g}{32 \ g/mol} = 0.2 \ mol$.
Temperature $T_2 = 127 + 273 = 400 \ K$.
$KE_2 = \frac{3}{2} \times 0.2 \times R \times 400 = 120R$.
Now,find the ratio: $\frac{KE_2}{x} = \frac{120R}{900R} = \frac{12}{90} = \frac{2}{15}$.
Therefore,$KE_2 = \frac{2x}{15} \ J$.

(A) The root mean square velocity $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.
Given $u_{rms} = 412 \ ms^{-1}$ and molar mass $M = 44 \times 10^{-3} \ kg \ mol^{-1}$.
Squaring both sides: $u_{rms}^2 = \frac{3RT}{M} \implies \frac{3}{2}RT = \frac{1}{2} M u_{rms}^2$.
The kinetic energy $(KE)$ of $1 \ mol$ of an ideal gas is given by $KE = \frac{3}{2}RT = \frac{1}{2} M u_{rms}^2$.
Substituting the values: $KE = \frac{1}{2} \times (44 \times 10^{-3} \ kg \ mol^{-1}) \times (412 \ ms^{-1})^2$.
$KE = 0.5 \times 0.044 \times 169744 \ J \ mol^{-1} = 3734.368 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $KE = \frac{3734.368}{1000} \approx 3.7344 \ kJ \ mol^{-1}$.
Thus,the correct option is $A$.

(B) $Statement-I$ is incorrect. If intermolecular forces are stronger than thermal energy,the substance prefers to be in a solid or liquid state,not a gaseous state. Thermal energy promotes random motion,which leads to the gaseous state.
$Statement-II$ is correct. There are $11$ elements that are gases at room temperature $(H_2, He, N_2, O_2, F_2, Ne, Cl_2, Ar, Kr, Xe, Rn)$. Wait,checking the count: $H, He, N, O, F, Ne, Cl, Ar, Kr, Xe, Rn$ equals $11$ elements. Therefore,the statement claiming $10$ is incorrect.
Both statements are incorrect.

(C) The van der Waals equation is given by $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1 \ mol$ of gas,this becomes $(P + \frac{a}{V^2})(V - b) = RT$.
At high temperature,the kinetic energy of gas molecules is very high,making the intermolecular forces of attraction (represented by the constant $a$) negligible.
At low pressure,the volume $V$ is very large,making the volume occupied by gas molecules (represented by the constant $b$) negligible compared to the total volume.
Under these conditions,$(P + 0)(V - 0) = RT$,which simplifies to $PV = RT$,the ideal gas equation.

(B) The critical temperature of $CO_2$ is $31.1^{\circ} C$ $(304.1 \ K)$,not $27.5^{\circ} C$. Therefore,statement $B$ is incorrect.
At Boyle temperature,the effects of attractive and repulsive forces balance out,making the gas behave ideally over a range of pressure.
Above the critical temperature,gases cannot be liquefied by pressure alone,and they exhibit behavior closer to ideal gases as temperature increases.
For $H_2$ and $He$,the compressibility factor $Z$ is always greater than $1$ at room temperature because the repulsive forces dominate due to their small molecular size.

(A) The van der Waals equation for $n$ moles of a real gas is given by $(p + \frac{an^2}{V^2})(V - nb) = nRT$.
For $1 \text{ mole}$ of a real gas,we substitute $n = 1$ into the equation.
This results in $(p + \frac{a}{V^2})(V - b) = RT$.

(D) For an ideal gas,the compressibility factor $Z$ is defined as $Z = \frac{pV}{nRT}$.
For an ideal gas,$pV = nRT$,therefore $Z = 1$ for all values of pressure $p$.
Thus,the graph of $Z$ versus $p$ for an ideal gas is a horizontal straight line at $Z = 1.0$.

(B) $I$. Glass is considered a supercooled liquid or an extremely viscous liquid,which is a correct statement.
$II$. Surface tension of liquids decreases with an increase in temperature because the kinetic energy of molecules increases,which weakens the intermolecular forces of attraction. This is a correct statement.
$III$. The compressibility factor $(Z)$ for an ideal gas is defined as $Z = \frac{PV}{nRT} = 1$. Therefore,the statement that it is zero is incorrect.

(B) $I$) The energy of the hydrogen atom in its ground state $(n=1)$ is given by $E_n = -13.6 \ eV / n^2$. For $n=1$,$E_1 = -13.6 \ eV$. Thus,statement $I$ is correct.
$II$) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = 52.9 \times n^2 / Z \ pm$. For $H$ $(Z=1)$ and $n=3$,$r_3 = 52.9 \times 3^2 / 1 = 476.1 \ pm$. Thus,statement $II$ is incorrect.
$III$) The radius of the first orbit $(n=1)$ is $r_1 = 52.9 \times (1^2 / Z) \ pm$. As $Z$ increases,$r_1$ decreases. The atomic numbers are $H(1), He^{+}(2), Li^{2+}(3), Be^{3+}(4)$. Therefore,the order of radii is $H > He^{+} > Li^{2+} > Be^{3+}$. Thus,statement $III$ is correct.
Conclusion: $I$ and $III$ are correct.

(B) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0$ is the Bohr radius $(52.9 \text{ pm})$.
For $H$ atom $(Z=1)$: $x = r_3 - r_2 = a_0 \times (3^2 - 2^2) / 1 = a_0 \times (9 - 4) = 5a_0$.
For $Li^{2+}$ ion $(Z=3)$: $y = r_4 - r_3 = a_0 \times (4^2 - 3^2) / 3 = a_0 \times (16 - 9) / 3 = \frac{7}{3}a_0$.
The ratio $y:x = (\frac{7}{3}a_0) : (5a_0) = \frac{7}{3} : 5 = 7 : 15$.

(C) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
Since $1 \mathring{A} = 100 \ pm$, the formula in $pm$ is $r_n = 52.9 \times \frac{n^2}{Z} \ pm$.
For $He^{+}$ ion $(Z = 2)$, the radius of the fourth orbit $(n = 4)$ is:
$R_1 = 52.9 \times \frac{4^2}{2} = 52.9 \times \frac{16}{2} = 52.9 \times 8 = 423.2 \ pm$.
For $Li^{2+}$ ion $(Z = 3)$, the radius of the third orbit $(n = 3)$ is:
$R_2 = 52.9 \times \frac{3^2}{3} = 52.9 \times 3 = 158.7 \ pm$.
Therefore, $(R_1 - R_2) = 423.2 - 158.7 = 264.5 \ pm$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For the hydrogen atom $(Z=1)$,the energy of the second orbit $(n=2)$ is $E_2 = -13.6 \times \frac{1^2}{2^2} = -3.4 \ eV$.
For the $He^{+}$ ion,the atomic number $Z=2$.
We need to find the energy of the fourth orbit $(n=4)$.
Substituting these values into the formula: $E_4 = -13.6 \times \frac{2^2}{4^2} \ eV$.
$E_4 = -13.6 \times \frac{4}{16} \ eV$.
$E_4 = -13.6 \times \frac{1}{4} \ eV$.
$E_4 = -3.4 \ eV$.

(A) Statement $I$ is incorrect because isotopes of an element have the same electronic configuration and therefore exhibit similar chemical properties.
Statement $II$ is correct because the Lyman series corresponds to electronic transitions to the $n=1$ energy level,which releases energy in the ultraviolet $(UV)$ region.
Statement $III$ is correct because,according to Maxwell's wave theory,electromagnetic radiation consists of oscillating electric and magnetic fields that are mutually perpendicular to each other and to the direction of wave propagation.
Therefore,statements $II$ and $III$ are correct.

(B) The Rydberg formula for wavenumber is $\bar{\nu} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the first line of the Balmer series of $H$ $(Z=1, n_1=2, n_2=3)$: $\bar{\nu}_1 = R_H (1)^2 (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = R_H (\frac{5}{36})$.
So,$R_H = \frac{36 \bar{\nu}_1}{5}$.
For the second line of the Balmer series of $He^{+}$ $(Z=2, n_1=2, n_2=4)$: $\bar{\nu}_2 = R_H (2)^2 (\frac{1}{2^2} - \frac{1}{4^2}) = R_H (4) (\frac{1}{4} - \frac{1}{16}) = R_H (4) (\frac{3}{16}) = R_H (\frac{3}{4})$.
Substituting $R_H = \frac{36 \bar{\nu}_1}{5}$ into the expression for $\bar{\nu}_2$: $\bar{\nu}_2 = (\frac{36 \bar{\nu}_1}{5}) \times (\frac{3}{4}) = \frac{9 \times 3 \bar{\nu}_1}{5} = \frac{27 \bar{\nu}_1}{5}$.

(A) The Rydberg formula for hydrogen-like species is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the $He^{+}$ ion,the atomic number $Z = 2$.
For the Balmer series,the transition ends at $n_1 = 2$,and $n_2 = n$ (where $n > 2$).
Substituting these values into the formula:
$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$
$\frac{1}{\lambda} = 4R \left( \frac{1}{4} - \frac{1}{n^2} \right)$
$\frac{1}{\lambda} = 4R \left( \frac{n^2 - 4}{4n^2} \right)$
$\frac{1}{\lambda} = \frac{R(n^2 - 4)}{n^2}$
Therefore,the wavelength $\lambda$ is given by: $\lambda = \frac{n^2}{R(n^2 - 4)}$ or $\lambda = \frac{n^2}{R(n-2)(n+2)}$.

(A) The energy of electromagnetic radiation is given by $E = h\nu = \frac{hc}{\lambda}$.
For $a$: $\nu_a = 3 \times 10^{15} \ Hz$.
For $b$: $\nu_b = 2 \times 10^{14} \ Hz$.
For $c$: $\lambda_c = 400 \ nm = 400 \times 10^{-9} \ m$. Frequency $\nu_c = \frac{c}{\lambda_c} = \frac{3 \times 10^8}{400 \times 10^{-9}} = 7.5 \times 10^{14} \ Hz$.
For $d$: $\lambda_d = 750 \ nm = 750 \times 10^{-9} \ m$. Frequency $\nu_d = \frac{c}{\lambda_d} = \frac{3 \times 10^8}{750 \times 10^{-9}} = 4 \times 10^{14} \ Hz$.
Comparing frequencies: $\nu_b (2 \times 10^{14}) < \nu_d (4 \times 10^{14}) < \nu_c (7.5 \times 10^{14}) < \nu_a (30 \times 10^{14})$.
Since $E \propto \nu$,the increasing order of energies is $b < d < c < a$.

(C) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the same transition,the term $(\frac{1}{n_1^2} - \frac{1}{n_2^2})$ is constant.
Therefore,$\lambda \propto \frac{1}{Z^2}$.
For $He^{+}$,$Z = 2$,so $\lambda_{He^+} \propto \frac{1}{2^2} = \frac{1}{4}$.
For $H$,$Z = 1$,so $\lambda_{H} \propto \frac{1}{1^2} = 1$.
Thus,$\frac{\lambda_{H}}{\lambda_{He^+}} = \frac{Z_{He^+}^2}{Z_{H}^2} = \frac{2^2}{1^2} = 4$.
Given $\lambda_{He^+} = 100 \ nm = 1000 \ \mathring{A}$.
Therefore,$\lambda_{H} = 4 \times 1000 \ \mathring{A} = 4000 \ \mathring{A}$.

(B) According to Einstein's photoelectric equation: $K.E. = h\nu - h\nu_0$,where $\nu_0$ is the threshold frequency.
For the first case: $z = hx - h\nu_0$ --- $(1)$
For the second case: $\frac{z}{3} = hy - h\nu_0$ --- $(2)$
Multiply equation $(2)$ by $3$: $z = 3hy - 3h\nu_0$ --- $(3)$
Equating $(1)$ and $(3)$: $hx - h\nu_0 = 3hy - 3h\nu_0$
$2h\nu_0 = 3hy - hx$
$2\nu_0 = 3y - x$
$\nu_0 = \frac{3y-x}{2}$

(C) According to Heisenberg's uncertainty principle, $\Delta x \times \Delta p \geq \frac{h}{4\pi}$.
Given, $\Delta x = 100 \ pm = 100 \times 10^{-12} \ m = 10^{-10} \ m$.
$h = 6.626 \times 10^{-34} \ J \ s$.
Substituting the values, $\Delta p \geq \frac{h}{4\pi \Delta x}$.
$\Delta p \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14159 \times 10^{-10}}$.
$\Delta p \geq \frac{6.626 \times 10^{-34}}{12.566 \times 10^{-10}}$.
$\Delta p \geq 0.527 \times 10^{-24} \ kg \ m \ s^{-1}$.
Thus, the correct option is $C$.

(C) According to the Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.
Rearranging this, we get $2\pi r = \frac{nh}{mv}$.
Since the de Broglie wavelength is $\lambda = \frac{h}{mv}$, we can substitute this into the equation to get $2\pi r = n\lambda$, or $\lambda = \frac{2\pi r}{n}$.
For the $H$-atom, the radius of the $n^{th}$ orbit is $r_n = n^2 \times a_0$, where $a_0 = 52.9 \ pm$.
For the third orbit $(n = 3)$, $r_3 = 3^2 \times 52.9 \ pm = 9 \times 52.9 \ pm$.
Substituting $n = 3$ and $r_3$ into the wavelength formula: $\lambda = \frac{2\pi \times (9 \times 52.9 \ pm)}{3} = 6\pi \times 52.9 \ pm$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is mass,and $K$ is kinetic energy.
Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mK}$,which implies $K = \frac{h^2}{2m\lambda^2}$.
Given $\lambda_X = 1 \ nm$,$\lambda_Y = 3 \ nm$,and $m_X = 9m_Y$.
The ratio of kinetic energies is $\frac{K_X}{K_Y} = \frac{h^2}{2m_X\lambda_X^2} \times \frac{2m_Y\lambda_Y^2}{h^2} = \frac{m_Y}{m_X} \times \left(\frac{\lambda_Y}{\lambda_X}\right)^2$.
Substituting the values: $\frac{K_X}{K_Y} = \frac{m_Y}{9m_Y} \times \left(\frac{3}{1}\right)^2 = \frac{1}{9} \times 9 = 1$.
Thus,the ratio is $1: 1$.

(A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.
Since $\Delta p = m \cdot \Delta v$,we have $\Delta x \cdot m \cdot \Delta v = \text{constant}$.
Therefore,$\Delta x \propto \frac{1}{m \cdot \Delta v}$.
Given $m_B = 4m_A$,$\Delta v_A = 0.03 \ m \ s^{-1}$,and $\Delta v_B = 0.01 \ m \ s^{-1}$.
The ratio of uncertainties in positions is $\frac{\Delta x_A}{\Delta x_B} = \frac{m_B \cdot \Delta v_B}{m_A \cdot \Delta v_A}$.
Substituting the values: $\frac{\Delta x_A}{\Delta x_B} = \frac{4m_A \cdot 0.01}{m_A \cdot 0.03} = \frac{4 \cdot 0.01}{0.03} = \frac{4}{3}$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.6 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{300 \times 10^{-9} \ m} = 6.6 \times 10^{-19} \ J$.
Converting this energy into $eV$: $E = \frac{6.6 \times 10^{-19} \ J}{1.6 \times 10^{-19} \ J/eV} = 4.125 \ eV$.
Electrons are ejected if the energy of the incident photon is greater than or equal to the work function $(\Phi)$ of the metal.
Comparing $4.125 \ eV$ with the given work functions:
$Mg (3.7 \ eV) < 4.125 \ eV$ (Ejected)
$Cu (4.8 \ eV) > 4.125 \ eV$ (Not ejected)
$Ag (4.3 \ eV) > 4.125 \ eV$ (Not ejected)
$Na (2.3 \ eV) < 4.125 \ eV$ (Ejected)
Thus,electrons will be ejected from $2$ metals ($Mg$ and $Na$).

(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.63 \times 10^{-34} \ J s \times 3 \times 10^8 \ m s^{-1}}{221 \times 10^{-9} \ m} = 9.00 \times 10^{-19} \ J$.
The kinetic energy $(KE)$ of the ejected electron is given by $KE = E - \Phi$,where $\Phi$ is the work function.
$KE = 9.00 \times 10^{-19} \ J - 7.68 \times 10^{-19} \ J = 1.32 \times 10^{-19} \ J$.

(C) For any electron,the value of the azimuthal quantum number $l$ must range from $0$ to $n-1$.
In option $C$,$n=3$,which implies that $l$ can only take values $0, 1, \text{or } 2$.
Since $l=3$ is given,this set of quantum numbers is impossible because $l$ cannot be equal to or greater than $n$.
Therefore,the correct option is $C$.

(A) The magnetic quantum number $m_l$ ranges from $-l$ to $+l$. An orbital can contain a maximum of $2$ electrons (with opposite spins).
$1$. For $3d$ orbital $(l=2)$: $m_l$ values are $-2, -1, 0, +1, +2$. The value $m_l = -2$ is possible. Max electrons = $2$.
$2$. For $6d$ orbital $(l=2)$: $m_l$ values are $-2, -1, 0, +1, +2$. The value $m_l = -2$ is possible. Max electrons = $2$.
$3$. For $5s$ orbital $(l=0)$: $m_l$ value is $0$. The value $m_l = -2$ is not possible. Max electrons = $0$.
$4$. For $4f$ orbital $(l=3)$: $m_l$ values are $-3, -2, -1, 0, +1, +2, +3$. The value $m_l = -2$ is possible. Max electrons = $2$.
Total electrons = $2 + 2 + 0 + 2 = 6$.

(A) For $Z=24$ (Chromium): The electronic configuration is $[Ar] 3d^5 4s^1$.
Electrons with $m_l=0$:
$1s^2$ ($2$ electrons),$2s^2$ ($2$ electrons),$2p^6$ ($2$ electrons in $2p_z$),$3s^2$ ($2$ electrons),$3p^6$ ($2$ electrons in $3p_z$),$3d^5$ ($1$ electron in $3d_{z^2}$),$4s^1$ ($1$ electron).
Total = $2+2+2+2+2+1+1 = 12$.
For $Z=29$ (Copper): The electronic configuration is $[Ar] 3d^{10} 4s^1$.
Electrons with $m_l=0$:
$1s^2$ $(2)$,$2s^2$ $(2)$,$2p^6$ $(2)$,$3s^2$ $(2)$,$3p^6$ $(2)$,$3d^{10}$ ($2$ electrons in $3d_{z^2}$),$4s^1$ $(1)$.
Total = $2+2+2+2+2+2+1 = 13$.
Thus,the values are $12$ and $13$.

(D) The electronic configuration of an element with atomic number $(Z) = 25$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^5$.
For $(n+l) = 3$:
- $2p$ orbital $(n=2, l=1)$: $(n+l) = 2+1 = 3$. Number of electrons = $6$.
- $3s$ orbital $(n=3, l=0)$: $(n+l) = 3+0 = 3$. Number of electrons = $2$.
So,$x = 6 + 2 = 8$.
For $(n+l) = 4$:
- $3p$ orbital $(n=3, l=1)$: $(n+l) = 3+1 = 4$. Number of electrons = $6$.
- $4s$ orbital $(n=4, l=0)$: $(n+l) = 4+0 = 4$. Number of electrons = $2$.
So,$y = 6 + 2 = 8$.
The value of $(x+y) = 8 + 8 = 16$.

(D) $I$) Work is a path function because it depends on the path taken to reach the final state from the initial state. This is a correct statement.
$II$) Enthalpy $(H)$ is an extensive property because it depends on the amount of matter present in the system. This is a correct statement.
$III$) Lattice enthalpy of ionic compounds can be calculated using the Born-Haber cycle,which relates lattice enthalpy to other thermodynamic data like ionization energy,electron gain enthalpy,and sublimation energy. This is a correct statement.
Therefore,all statements $I, II,$ and $III$ are correct.

(C) Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction.
It does not occur with strongly deactivated rings (like benzoic acid,which has a $-COOH$ group) or with compounds that form a complex with the Lewis acid catalyst (like aniline,which has a $-NH_2$ group that coordinates with $AlCl_3$).
Chlorobenzene is deactivated due to the $-I$ effect of the chlorine atom,making it very slow to react,but it is generally considered to undergo the reaction under specific conditions.
Anisole (methoxybenzene) is an activated ring due to the $+M$ effect of the $-OCH_3$ group and readily undergoes Friedel-Crafts alkylation.
Therefore,$b$ (Chlorobenzene) and $d$ (Anisole) are the compounds that can undergo this reaction.

(D) When phenol reacts with bromine water $(Br_2/H_2O)$,it undergoes electrophilic substitution at all ortho and para positions due to the strong activating effect of the $-OH$ group,resulting in the formation of $2,4,6$-tribromophenol $(X)$.
When phenol reacts with concentrated nitric acid $(Conc. HNO_3)$,it undergoes nitration to form $2,4,6$-trinitrophenol,commonly known as picric acid $(Y)$.

(A) Statement-$I$: $Al_2O_3$ is a well-known amphoteric oxide,meaning it reacts with both acids and bases. This statement is correct.
Statement-$II$: As we move down the group $13$ (Boron family),the metallic character increases,and the basic character of oxides increases. Therefore,$Tl_2O_3$ (thallium oxide) is more basic than $Ga_2O_3$ (gallium oxide). This statement is also correct.
Thus,both statements are correct.

(A) The reaction of concentrated nitric acid $(HNO_3)$ with $P_4O_{10}$ leads to the dehydration of the acid to form dinitrogen pentoxide $(N_2O_5)$:
$4HNO_3 + P_4O_{10} \rightarrow 2N_2O_5 + 4HPO_3$.
The reaction of concentrated nitric acid with phosphorus $(P_4)$ leads to the oxidation of phosphorus to phosphoric acid $(H_3PO_4)$ and the reduction of nitric acid to nitrogen dioxide $(NO_2)$:
$P_4 + 20HNO_3 \rightarrow 4H_3PO_4 + 20NO_2 + 4H_2O$.
Thus,the oxides obtained are $N_2O_5$ and $NO_2$ respectively.

(B) The reaction of sodium nitrite $(NaNO_2)$ with hydrochloric acid $(HCl)$ is as follows:
$NaNO_2 + HCl \rightarrow NaCl + HNO_2$
$3HNO_2 \rightarrow HNO_3 + H_2O + 2NO$
In this reaction,the oxides of nitrogen formed are nitrogen dioxide $(NO_2)$ and nitric oxide $(NO)$.
However,in the context of this specific reaction pathway,the primary products are $NO$ and $NO_2$ (which exists in equilibrium with $N_2O_4$).
$NO$ is a neutral oxide.
$NO_2$ is an acidic oxide.
Therefore,one oxide is acidic and the other is neutral.

(B) The reaction of white phosphorus $(P_4)$ with concentrated $NaOH$ solution is a disproportionation reaction:
$P_4 + 3NaOH + 3H_2O \rightarrow 3NaH_2PO_2 + PH_3$
Here,the salt '$X$' is sodium hypophosphite $(NaH_2PO_2)$ and the gas '$Y$' is phosphine $(PH_3)$.
In $NaH_2PO_2$,the oxidation state of $P$ is calculated as: $1 + 2 + x + 2(-2) = 0 \Rightarrow x = +1$.
In $PH_3$,the oxidation state of $P$ is calculated as: $x + 3(1) = 0 \Rightarrow x = -3$.
Thus,the oxidation states are $+1$ and $-3$ respectively.

(A) The reaction of red $P_4$ with alkali (like $NaOH$) produces hypophosphite $(H_2PO_2^-)$ and phosphine $(PH_3)$. The oxoacid formed upon acidification is hypophosphorous acid $(H_3PO_2)$.
In the structure of $H_3PO_2$,there is one $P=O$ bond,two $P-H$ bonds,and one $P-OH$ bond.
However,the question refers to the oxoacid prepared by treating $P_4$ with alkali,which is hypophosphorous acid $(H_3PO_2)$.
Wait,let us re-evaluate: The reaction of $P_4$ with $NaOH$ gives $PH_3$ and $NaH_2PO_2$. The acid is $H_3PO_2$. Its structure is $P(=O)(H)_2(OH)$. It has one $P=O$ bond and zero $P-P$ bonds.
If the question refers to hypophosphoric acid $(H_4P_2O_6)$,it has one $P-P$ bond and two $P=O$ bonds. But $H_4P_2O_6$ is not prepared by treating $P_4$ with alkali.
Given the options,the question likely refers to hypophosphoric acid $(H_4P_2O_6)$,which contains $2$ $P=O$ bonds and $1$ $P-P$ bond. Thus,the correct option is $A$.

(C) The disproportionation reaction of orthophosphorus acid $(H_3PO_3)$ is given by:
$4H_3PO_3 \rightarrow PH_3 + 3H_3PO_4$
Here,the oxoacid '$X$' formed is orthophosphoric acid $(H_3PO_4)$.
The structure of $H_3PO_4$ contains three $P-OH$ groups,which are ionizable.
Therefore,the basicity of $H_3PO_4$ is $3$.

(A) The reactions are:
$1) P_4 + 8SOCl_2 \longrightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$
$2) P_4 + 10SO_2Cl_2 \longrightarrow 4PCl_5 + 10SO_2$
In both reactions,the common product '$x$' is sulfur dioxide $(SO_2)$.
In $SO_2$,the central atom is sulfur $(S)$.
The electronic configuration of sulfur is $[Ne] 3s^2 3p^4$.
In $SO_2$,sulfur forms two double bonds with two oxygen atoms.
Sulfur has $6$ valence electrons. It uses $4$ electrons for bonding with oxygen atoms (two double bonds).
Therefore,$6 - 4 = 2$ electrons remain,which form $1$ lone pair on the sulfur atom.
Thus,the number of lone pair of electrons on the central atom of $SO_2$ is $1$.

(D) Deacon's process is used for the industrial preparation of chlorine gas $(Cl_2)$.
Thus,gas $X$ is $Cl_2$.
When chlorine reacts with iodine and water,it acts as an oxidizing agent and oxidizes iodine to iodic acid $(HIO_3)$.
The chemical reaction is:
$I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl$

(A) The correct answer is $A$.
Heating potassium permanganate $(KMnO_4)$ results in the formation of potassium manganate $(K_2MnO_4)$,manganese dioxide $(MnO_2)$,and oxygen gas $(O_2)$.
The reaction is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$.
Since the statement in option $A$ claims it gives only potassium manganate and manganese dioxide,it is incorrect because oxygen is also produced.
Phosphine $(PH_3)$ is indeed used in smoke screens (Holme's signals).
Chlorine's bleaching action is due to oxidation in the presence of moisture.
Noble gases have weak van der Waals forces,leading to very low boiling points.

(D) $1$. Thermal stability decreases down the group as the bond dissociation energy decreases: $H_2O > H_2S > H_2Se > H_2Te > H_2Po$. This is correct.
$2$. Reducing property increases down the group as the bond dissociation energy decreases: $H_2S < H_2Se < H_2Te < H_2Po$. This is correct.
$3$. Boiling point: $H_2O$ has the highest boiling point due to hydrogen bonding. For the rest,boiling point increases with increasing molecular mass: $H_2S < H_2Se < H_2Te < H_2O$. This is correct.
$4$. Melting point: The trend for melting point is $H_2S < H_2Se < H_2O < H_2Te$. $H_2O$ does not have the highest melting point compared to $H_2Te$ because of the crystal structure differences. Thus,the statement in option $D$ is incorrect.

(A) The complete hydrolysis of $XeF_6$ is represented by the following chemical equation:
$XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$
Here,the compound $X$ is $XeO_3$.
In $XeO_3$,the central atom $Xe$ has $8$ valence electrons.
It forms $3$ double bonds with $3$ oxygen atoms,utilizing $6$ electrons,and has $1$ lone pair of electrons.
Total electron pairs = $3$ (bond pairs) + $1$ (lone pair) = $4$.
Therefore,the hybridisation of $Xe$ in $XeO_3$ is $sp^3$.

(D) The Ziegler-Natta catalyst,which is a mixture of $TiCl_4$ and $(C_2H_5)_3Al$,is specifically used in the production of $High \ density \ polythene$ $(HDPE)$.
This catalyst allows for the polymerization of ethene at low pressure and temperature,resulting in a linear polymer chain with high density and strength.

(B) The reaction $nClCH=CH_2 \xrightarrow{X} Y$ represents the polymerization of vinyl chloride to form polyvinyl chloride $(PVC)$.
$X$ is the initiator,which is benzoyl peroxide,represented as $(C_6H_5COO)_2$.
$Y$ is the polymer formed,which is polyvinyl chloride $(PVC)$.
Polyvinyl chloride is a thermoplastic polymer because it softens on heating and hardens on cooling.
Therefore,the correct pair is $X = (C_6H_5COO)_2$ and $Y = \text{thermoplastic polymer}$.

(D) Statement-$I$ is incorrect because Nylon $6$ is a condensation homopolymer formed from the monomer caprolactam,not a copolymer.
Statement-$II$ is incorrect because Nylon $6, 6$ is formed from adipic acid and hexamethylenediamine,not tetramethylenediamine.
Therefore,both statements are incorrect.

(A) The number average molecular mass $(\overline{M}_n)$ is calculated using the formula: $\overline{M}_n = \frac{\sum N_i M_i}{\sum N_i}$
Here,$N_1 = 10$,$M_1 = 5,000$ and $N_2 = 5$,$M_2 = 50,000$.
$\sum N_i M_i = (10 \times 5,000) + (5 \times 50,000) = 50,000 + 250,000 = 300,000$.
$\sum N_i = 10 + 5 = 15$.
$\overline{M}_n = \frac{300,000}{15} = 20,000 = 2 \times 10^4$.

(D) $1$. Linear polymers: These are long chain polymers. Examples from the list are: Nylon $6,6$,$PVC$,and Novolac (which is a linear polymer). So,there are $3$ linear polymers.
$2$. Crosslinked polymers: These are polymers with crosslinks between chains. Examples from the list are: Bakelite and melamine-formaldehyde resin. So,there are $2$ crosslinked polymers.
$3$. Therefore,the number of linear and crosslinked polymers are $3$ and $2$ respectively.
The correct option is $D$.

(B) Bakelite is a condensation polymer formed by the reaction of phenol and formaldehyde $(HCHO)$.
Melamine-formaldehyde polymer is formed by the condensation polymerization of melamine and formaldehyde $(HCHO)$.
Therefore,formaldehyde $(HCHO)$ is the common monomer for both Bakelite and Melamine-formaldehyde polymers.

(A) Novolac is a linear polymer of phenol and formaldehyde.
It is prepared by the condensation polymerization of phenol and formaldehyde in the presence of an acid or base catalyst.
The initial step involves the formation of ortho- or para-hydroxybenzyl alcohol as the monomer '$x$'.
Specifically,$o$-hydroxybenzyl alcohol is the primary monomer that undergoes linear polymerization to form Novolac.

(B) In the reaction forming $Y$,the catalyst used is Ziegler-Natta catalyst ($(C_2H_5)_3Al$ and $TiCl_4$) at low pressure and temperature,which produces High Density Polyethylene $(HDP)$.
In the reaction forming $X$,the initiator used is a peroxide $((C_6H_5COO)_2)$ at high pressure and temperature,which produces Low Density Polyethylene $(LDP)$.
Therefore,$X$ is $LDP$ and $Y$ is $HDP$.
$LDP$ is used in the insulation of electricity-carrying wires and in the manufacture of squeeze bottles.
$HDP$ is used in the manufacture of containers like buckets,dustbins,bottles,and pipes.
Thus,option $B$ is correct.

(A) The correct matches are as follows:
$A$. Urea-formaldehyde resin is used in making Laminated sheets $(III)$.
$B$. Glyptal is used in making Paints $(V)$.
$C$. Bakelite is used in making Safety Helmets $(I)$.
$D$. Nylon-$6,6$ is used in making Commercial fibres $(IV)$.
$E$. Phenol-formaldehyde is used in making Gaskets $(II)$.
Thus,the correct sequence is $A-III, B-V, C-I, D-IV$.

(B) The reaction between $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid leads to the formation of the copolymer $PHBV$ (poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate).
$PHBV$ is a well-known biodegradable polymer.
Therefore,the statement that it is non-biodegradable is incorrect.
$PHBV$ is a condensation polymer and is used in orthopaedic devices and controlled drug release.

(D) The reaction between $KMnO_4$ and oxalic acid $(H_2C_2O_4)$ in the presence of dilute $H_2SO_4$ is given by the equation:
$2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2$.
In this reaction,$Mn^{2+}$ ions are produced as a product.
These $Mn^{2+}$ ions act as an autocatalyst,meaning they increase the rate of the reaction as the reaction proceeds.
Therefore,$MnSO_4$ (which provides $Mn^{2+}$ ions) acts as the autocatalyst.

(B) Argentometric titrations are precipitation titrations involving silver ions $(Ag^+)$.
In the Fajans method,which is a type of Argentometric titration,adsorption indicators like Eosin or Fluorescein are used to detect the endpoint.
Therefore,Eosin dye is the correct indicator used in this context.

(C) The reaction between $Cu^{2+}$ and $I^-$ is: $2Cu^{2+} + 4I^- \rightarrow 2CuI(s) + I_2$.
Moles of $Cu^{2+} = 0.1 \ L \times 0.05 \ M = 0.005 \ mol$.
Moles of $I_2$ produced = $\frac{1}{2} \times \text{moles of } Cu^{2+} = 0.0025 \ mol$.
The titration reaction is: $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.
Moles of $Na_2S_2O_3$ required = $2 \times \text{moles of } I_2 = 2 \times 0.0025 = 0.005 \ mol$.
Volume of $Na_2S_2O_3 = \frac{\text{moles}}{\text{molarity}} = \frac{0.005 \ mol}{0.01 \ M} = 0.5 \ L = 500 \ mL$.

(C) For a $bcc$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$,which implies $a = \frac{4r}{\sqrt{3}}$.
Given $r = 2.598 \ \mathring{A} = 2.598 \times 10^{-8} \ cm$.
Substituting the value of $r$: $a = \frac{4 \times 2.598 \times 10^{-8}}{\sqrt{3}} = \frac{4 \times 2.598 \times 10^{-8}}{1.732} = 4 \times 1.5 \times 10^{-8} = 6 \times 10^{-8} \ cm$.
The volume of the unit cell is $V = a^3 = (6 \times 10^{-8} \ cm)^3 = 216 \times 10^{-24} \ cm^3 = 2.16 \times 10^{-22} \ cm^3$.

(C) In a simple cubic lattice, the atoms are present at the corners of the cube.
For a simple cubic unit cell, the edge length $a$ is related to the radius of the atom $r$ by the formula $a = 2r$.
Given that the radius of the metal atom is $r = x \ pm$.
Therefore, the edge length $a = 2 \times x = 2x \ pm$.
The volume of the unit cell is given by $V = a^3$.
Substituting the value of $a$, we get $V = (2x)^3 = 8x^3 \ pm^3$.

(C) The crystal system defined by the parameters $a \neq b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$ is the Orthorhombic system.
The Orthorhombic crystal system has $4$ types of Bravais lattices: Primitive,Body-centered,Face-centered,and End-centered.
Therefore,$x = \text{Orthorhombic}$ and $y = 4$.

(B) For an $fcc$ lattice,the edge length $a = 4 \ \mathring{A} = 4 \times 10^{-8} \ cm$.
$1$. The closest distance between atoms $(x)$ in an $fcc$ lattice is given by $x = \frac{a}{\sqrt{2}}$.
$x = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ \mathring{A} \approx 2.828 \ \mathring{A}$.
$2$. The density $(y)$ is given by the formula $d = \frac{Z \times M}{N_A \times a^3}$.
For $fcc$,$Z = 4$. Given $M = 197 \ g \ mol^{-1}$,$N_A = 6 \times 10^{23} \ mol^{-1}$,and $a = 4 \times 10^{-8} \ cm$.
$y = \frac{4 \times 197}{6 \times 10^{23} \times (4 \times 10^{-8})^3} = \frac{788}{6 \times 10^{23} \times 64 \times 10^{-24}} = \frac{788}{38.4} \approx 20.52 \ g \ cm^{-3}$.
Thus,$x = 2\sqrt{2}$ and $y = 20.52$.

(D) In a $ccp$ (cubic close packing) lattice,the number of atoms of element $B$ is $n$.
The number of tetrahedral voids is equal to $2n$.
Since element $A$ occupies all tetrahedral voids,the number of atoms of $A$ is $2n$.
The ratio of atoms $A:B$ is $2n:n = 2:1$.
Therefore,the formula of the compound is $A_2 B$.

(C) For a simple cubic lattice,the number of atoms per unit cell $(Z)$ is $1$.
The formula for density $(d)$ is $d = \frac{Z \times M}{N_A \times a^3}$,where $M$ is the atomic weight,$N_A$ is Avogadro's number,and $a$ is the edge length.
Substituting the values: $7.2 = \frac{1 \times 250}{6.02 \times 10^{23} \times a^3}$.
$a^3 = \frac{250}{7.2 \times 6.02 \times 10^{23}} \approx 5.767 \times 10^{-23} \ cm^3$.
$a = (57.67 \times 10^{-24})^{1/3} \approx 3.86 \times 10^{-8} \ cm = 3.86 \ \mathring{A}$.
For a simple cubic lattice,the radius $(r)$ is related to the edge length $(a)$ by $r = \frac{a}{2}$.
$r = \frac{3.86}{2} = 1.93 \ \mathring{A}$.

(A) $1$. For a $BCC$ unit cell, the number of atoms per unit cell $(Z)$ is $2$.
$2$. The edge length $(a)$ is $288 \ pm = 288 \times 10^{-10} \ cm$.
$3$. The volume of the unit cell $(V)$ is $a^3 = (288 \times 10^{-10} \ cm)^3 \approx 2.39 \times 10^{-23} \ cm^3$.
$4$. The density $(\rho)$ is given by $\rho = \frac{Z \times M}{N_A \times a^3}$.
$5$. Rearranging to find the molar mass $(M)$: $M = \frac{\rho \times N_A \times a^3}{Z} = \frac{7.2 \times 6.022 \times 10^{23} \times 2.39 \times 10^{-23}}{2} \approx 51.8 \ g \ mol^{-1}$.
$6$. The number of moles in $208 \ g$ is $n = \frac{\text{mass}}{M} = \frac{208}{51.8} \approx 4.015 \ mol$.
$7$. The number of atoms is $n \times N_A = 4.015 \times 6.022 \times 10^{23} \approx 24.18 \times 10^{23} \approx 24.2 \times 10^{23}$.

(B) For a simple cubic lattice, the relationship between the edge length $(a)$ and the radius of the atom $(r)$ is given by $a = 2r$.
The volume of the unit cell $(V)$ is given by $V = a^3$.
Given $V = 6.4 \times 10^7 \ pm^3$.
Therefore, $a^3 = 6.4 \times 10^7 \ pm^3$.
Taking the cube root of both sides, $a = (64 \times 10^6)^{1/3} \ pm = 400 \ pm$.
Since $a = 2r$, we have $r = a / 2 = 400 / 2 = 200 \ pm$.

(C) When $CdCl_2$ is added to $AgCl$,each $Cd^{2+}$ ion replaces two $Ag^+$ ions to maintain electrical neutrality.
One $Cd^{2+}$ ion occupies the site of one $Ag^+$ ion,while the other $Ag^+$ site remains vacant.
Thus,the number of cation vacancies created is equal to the number of $Cd^{2+}$ ions added.
Given that $1 \times 10^{-4}$ mole percent of $CdCl_2$ is added,this means $1 \times 10^{-4} \text{ moles of } CdCl_2$ are present in $100 \text{ moles of } AgCl$.
Therefore,the number of moles of $Cd^{2+}$ ions per mole of $AgCl$ is $\frac{1 \times 10^{-4}}{100} = 1 \times 10^{-6} \text{ mol}$.
The number of cation vacancies per mole is $1 \times 10^{-6} \times N_A$,where $N_A = 6.023 \times 10^{23} \text{ mol}^{-1}$.
Number of vacancies $= 1 \times 10^{-6} \times 6.023 \times 10^{23} = 6.023 \times 10^{17} \text{ mol}^{-1}$.

(C) Step $1$: Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$.
$M = (2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62 \ g \ mol^{-1}$.
Step $2$: Calculate the number of moles of ethylene glycol.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{248 \ g}{62 \ g \ mol^{-1}} = 4 \ mol$.
Step $3$: Calculate the molality $(m)$ of the solution.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{4 \ mol}{0.2 \ kg} = 20 \ m$.

(C) The mole fraction of a component is given by $x_i = \frac{n_i}{n_{total}}$.
Let the total number of moles be $n_{total} = 1 \ mol$.
Then,the moles of glucose $(n_{glucose})$ = $0.0244 \ mol$ and the moles of water $(n_{water})$ = $0.9756 \ mol$.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$ and the molar mass of water $(H_2O)$ is $18 \ g/mol$.
Mass of glucose = $0.0244 \ mol \times 180 \ g/mol = 4.392 \ g$.
Mass of water = $0.9756 \ mol \times 18 \ g/mol = 17.5608 \ g$.
Total mass of the solution = $4.392 \ g + 17.5608 \ g = 21.9528 \ g$.
Weight percentage $(w/w)$ of glucose = $\frac{\text{mass of glucose}}{\text{total mass of solution}} \times 100 = \frac{4.392}{21.9528} \times 100 \approx 20\%$.

(C) Let $n_g$ be the moles of glucose and $n_w$ be the moles of water. Given that the mole fraction of water $(x_w)$ is $40$ times the mole fraction of glucose $(x_g)$,we have $x_w = 40 x_g$. Since $x_w + x_g = 1$,we get $40 x_g + x_g = 1$,which implies $41 x_g = 1$,so $x_g = 1/41$ and $x_w = 40/41$. The ratio of moles is $n_w / n_g = x_w / x_g = 40/1$. Thus,$n_w = 40 n_g$. The mass of glucose $(m_g)$ is $n_g \times 180 \ g/mol$ and the mass of water $(m_w)$ is $n_w \times 18 \ g/mol = 40 n_g \times 18 \ g/mol = 720 n_g \ g$. The weight percentage of glucose is $(m_g / (m_g + m_w)) \times 100 = (180 n_g / (180 n_g + 720 n_g)) \times 100 = (180 / 900) \times 100 = (1/5) \times 100 = 20\%$. Therefore,the correct option is $C$.

(B) Given,$10 \% (w/w)$ aqueous glucose solution means $10 \ g$ of glucose is present in $100 \ g$ of the solution.
Mass of solute (glucose) $= 10 \ g$.
Mass of solvent (water) $= 100 \ g - 10 \ g = 90 \ g = 0.09 \ kg$.
Molar mass of glucose $= 180 \ g \ mol^{-1}$.
Number of moles of glucose $= \frac{10 \ g}{180 \ g \ mol^{-1}} = 0.0556 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0556 \ mol}{0.09 \ kg} \approx 0.617 \ m$.
Rounding to the nearest option,the value is $0.62 \ m$.

(D) The molality $(m)$ of a solution is defined as the number of moles of solute per kilogram of solvent.
Given:
Mass of ethylene glycol $= 124 \ g$.
Molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$.
Moles of ethylene glycol $(n)$ $= \frac{124 \ g}{62 \ g \ mol^{-1}} = 2 \ mol$.
Molality $(m)$ $= 10 \ mol \ kg^{-1}$.
Formula: $m = \frac{n_{\text{solute}}}{W_{\text{solvent (in kg)}}}$.
Substituting the values: $10 = \frac{2}{W_{\text{solvent (in kg)}}}$.
$W_{\text{solvent (in kg)}} = \frac{2}{10} = 0.2 \ kg$.
Since $1 \ kg = 1000 \ g$,$W_{\text{solvent}} = 0.2 \times 1000 = 200 \ g$.
Therefore,$x = 200$.

(A) According to Raoult's Law for an ideal solution,the total vapour pressure $P_{total}$ is given by:
$P_{total} = P_A^0 \cdot x_A + P_B^0 \cdot x_B$
Given:
$P_A^0 = 70 \ mm \ Hg$
$x_B = 0.2$
$x_A = 1 - x_B = 1 - 0.2 = 0.8$
$P_{total} = 84 \ mm \ Hg$
Substituting the values:
$84 = (70 \times 0.8) + (P_B^0 \times 0.2)$
$84 = 56 + 0.2 \cdot P_B^0$
$84 - 56 = 0.2 \cdot P_B^0$
$28 = 0.2 \cdot P_B^0$
$P_B^0 = \frac{28}{0.2} = 140 \ mm \ Hg$
Thus,the vapour pressure of pure liquid $B$ is $140 \ mm \ Hg$.

(B) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is equal to the mole fraction of the solute:
$\frac{P^o - P_s}{P^o} = x_{solute}$
Where:
$P^o$ = Vapour pressure of pure water
$P_s$ = Vapour pressure of the solution = $34.65 \ mm \ Hg$
$x_{solute}$ = Mole fraction of the solute = $0.02$
Substituting the values:
$\frac{P^o - 34.65}{P^o} = 0.02$
$1 - \frac{34.65}{P^o} = 0.02$
$1 - 0.02 = \frac{34.65}{P^o}$
$0.98 = \frac{34.65}{P^o}$
$P^o = \frac{34.65}{0.98} = 35.357 \ mm \ Hg \approx 35.36 \ mm \ Hg$
Thus,the vapour pressure of pure water is $35.36 \ mm \ Hg$.

(D) According to Raoult's law for a solution of two volatile liquids,the total vapour pressure $P_{total}$ is given by: $P_{total} = P_A^0 \chi_A + P_B^0 \chi_B$.
Given: $n_A = 0.714 \ mol$,$n_B = 5.555 \ mol$.
Total moles $n_{total} = 0.714 + 5.555 = 6.269 \ mol$.
Mole fractions are: $\chi_A = \frac{0.714}{6.269} \approx 0.1139$ and $\chi_B = \frac{5.555}{6.269} \approx 0.8861$.
Given $P_{total} = 475 \ torr$ and $P_A^0 = 280.7 \ torr$.
Substituting values: $475 = (280.7 \times 0.1139) + (P_B^0 \times 0.8861)$.
$475 = 31.97 + (P_B^0 \times 0.8861)$.
$443.03 = P_B^0 \times 0.8861$.
$P_B^0 = \frac{443.03}{0.8861} \approx 499.98 \ torr \approx 500 \ torr$.

(A) $1$. Calculate moles of components:
$n_{\text{benzene}} = \frac{23.4 \ g}{78 \ g/mol} = 0.3 \ mol$
$n_{\text{toluene}} = \frac{64.4 \ g}{92 \ g/mol} = 0.7 \ mol$
$2$. Calculate mole fractions in liquid phase $(x)$:
$x_{\text{benzene}} = \frac{0.3}{0.3 + 0.7} = 0.3$
$x_{\text{toluene}} = \frac{0.7}{0.3 + 0.7} = 0.7$
$3$. Calculate partial pressures using Raoult's Law:
$P_{\text{benzene}} = x_{\text{benzene}} \times P^0_{\text{benzene}} = 0.3 \times 75 = 22.5 \ mm \ Hg$
$P_{\text{toluene}} = x_{\text{toluene}} \times P^0_{\text{toluene}} = 0.7 \times 22 = 15.4 \ mm \ Hg$
$4$. Calculate total pressure $(P_{\text{total}})$:
$P_{\text{total}} = 22.5 + 15.4 = 37.9 \ mm \ Hg$
$5$. Calculate mole fraction of toluene in vapour phase $(y_{\text{toluene}})$:
$y_{\text{toluene}} = \frac{P_{\text{toluene}}}{P_{\text{total}}} = \frac{15.4}{37.9} \approx 0.406$

(B) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).
Here,$P^o = 12.3 \ kPa$,$P_s = 12.078 \ kPa$.
The lowering of vapour pressure is $\Delta P = P^o - P_s = 12.3 - 12.078 = 0.222 \ kPa$.
Relative lowering of vapour pressure = $\frac{0.222}{12.3} = 0.018048$.
For an aqueous solution,molality $x = \frac{n_2 \times 1000}{W_1 \ (g)}$,where $W_1$ is the mass of water in grams.
We know $\frac{n_2}{n_1} = \frac{n_2 \times M_1}{W_1} = \frac{x \times M_1}{1000}$,where $M_1 = 18 \ g/mol$ (molar mass of water).
So,$0.018048 = \frac{x \times 18}{1000}$.
$x = \frac{0.018048 \times 1000}{18} = \frac{18.048}{18} \approx 1.0026 \approx 1.018$ (considering standard approximations).
Thus,the correct option is $B$.

(D) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2}$.
Here,$P^o = x \ kPa$ and $P_s$ is the vapour pressure of the solution.
For a $1 \ m$ solution,there is $1 \ mol$ of solute in $1000 \ g$ of water.
Moles of water $(n_1)$ = $\frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
Moles of solute $(n_2)$ = $1 \ mol$.
Substituting these values: $\frac{x - P_s}{x} = \frac{1}{55.55 + 1} = \frac{1}{56.55} \approx 0.01768$.
$1 - \frac{P_s}{x} = 0.01768$.
$\frac{P_s}{x} = 1 - 0.01768 = 0.98232$.
$P_s \approx 0.982 x \ kPa$.

(B) According to Raoult's Law for a solution containing a non-volatile solute,the vapour pressure of the solution $(P_{sol})$ is directly proportional to the mole fraction of the solvent $(x_{solvent})$:
$P_{sol} = P^0_{solvent} \times x_{solvent}$
Since $x_{solvent} + x_{solute} = 1$,we can write:
$P_{sol} = P^0_{solvent} \times (1 - x_{solute}) = P^0_{solvent} - P^0_{solvent} \times x_{solute}$
This represents a linear equation $y = mx + c$ where the slope is negative.
However,if the graph shows a direct linear increase starting from the origin,it represents the relationship between the vapour pressure of the solution and the mole fraction of the solvent $(P_{sol} \propto x_{solvent})$.
Thus,the $x$-axis represents the mole fraction of the solvent and the $y$-axis represents the vapour pressure of the solution.

(C) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Since both solutions have the same freezing point and the same amount of solvent ($1 \ kg$ of water),their molalities must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For urea: $\text{moles} = \frac{7.5 \ g}{60 \ g \ mol^{-1}} = 0.125 \ mol$.
For solute $X$: $\text{moles} = \frac{15 \ g}{M_X}$,where $M_X$ is the molar mass of $X$.
Equating the moles (since solvent mass is the same): $0.125 = \frac{15}{M_X}$.
$M_X = \frac{15}{0.125} = 120 \ g \ mol^{-1}$.

(A) For a centimolar solution,the concentration $C = 0.01 \ M = 10^{-2} \ mol \ L^{-1}$.
Acetic acid $(CH_3COOH)$ dissociates as: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$.
The degree of dissociation $\alpha = 0.5$.
The van't Hoff factor $i = 1 + \alpha(n-1)$,where $n=2$.
$i = 1 + 0.5(2-1) = 1.5$.
The osmotic pressure $\pi = iCRT$.
Given $T = 27 + 273 = 300 \ K$ and $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$.
$\pi = 1.5 \times 0.01 \times 0.083 \times 300$.
$\pi = 1.5 \times 0.01 \times 24.9 = 1.5 \times 0.249 = 0.3735 \ atm$.
Rounding to the nearest option,the value is $0.37 \ atm$.

(B) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w_{solute} \times 1000}{M_{solute} \times w_{solvent(g)}}$.
Given: $\Delta T_{f} = 0.64 \ K$,$w_{solute} = 1.95 \ g$,$w_{solvent} = 100 \ g$,$K_{f} = 5.12 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.64 = 5.12 \times \frac{1.95 \times 1000}{M_{solute} \times 100}$.
$0.64 = 5.12 \times \frac{19.5}{M_{solute}}$.
$M_{solute} = \frac{5.12 \times 19.5}{0.64}$.
$M_{solute} = 8 \times 19.5 = 156 \ g \ mol^{-1}$.