MHT CET 2023 Chemistry Question Paper with Answer and Solution

(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi_0$,where $\phi_0 = \frac{hc}{\lambda_0}$ and $\lambda_0$ is the threshold wavelength.
For the first case: $e(4.8) = \frac{hc}{\lambda} - \phi_0 \quad ... (i)$
For the second case: $e(1.6) = \frac{hc}{2\lambda} - \phi_0 \quad ... (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{4.8}{1.6} = \frac{\frac{hc}{\lambda} - \phi_0}{\frac{hc}{2\lambda} - \phi_0}$
$3 = \frac{\frac{hc}{\lambda} - \phi_0}{\frac{hc}{2\lambda} - \phi_0}$
$3 \left( \frac{hc}{2\lambda} - \phi_0 \right) = \frac{hc}{\lambda} - \phi_0$
$\frac{3hc}{2\lambda} - 3\phi_0 = \frac{hc}{\lambda} - \phi_0$
$\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = 2\phi_0$
$\frac{hc}{2\lambda} = 2\phi_0 \implies \phi_0 = \frac{hc}{4\lambda}$
Since $\phi_0 = \frac{hc}{\lambda_0}$,we have $\frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$.
Therefore,the threshold wavelength $\lambda_0 = 4\lambda$.

(B) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.
For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(1)$
For the second case: $e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$4 = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$
$4 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$4(\frac{1}{2\lambda} - \frac{1}{\lambda_0}) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{2}{\lambda} - \frac{4}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{2}{\lambda} - \frac{1}{\lambda} = \frac{4}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{\lambda} = \frac{3}{\lambda_0}$
$\lambda_0 = 3\lambda$.

(D) Consider a small radial element of length $dr$ at a distance $r$ from the centre of the disc.
As the disc rotates,this element moves with a linear velocity $v = r\omega$ in the magnetic field $B$.
The motional e.m.f. $de$ induced across this small element is given by $de = Bv dr = B(r\omega) dr$.
To find the total induced e.m.f. $e$ between the centre $(r=0)$ and the rim $(r=R)$,we integrate the expression:
$e = \int_{0}^{R} B\omega r dr$
$e = B\omega \int_{0}^{R} r dr$
$e = B\omega \left[ \frac{r^2}{2} \right]_{0}^{R}$
$e = \frac{B\omega R^2}{2}$
The magnitude of the induced e.m.f. is $|e| = \frac{B\omega R^2}{2}$.

(D) The magnetic field $B$ at the center of a circular loop of radius $R_1$ carrying current $I_1$ is given by $B = \frac{\mu_0 I_1}{2 R_1}$.
Since $R_1$ > $R_2$, we assume the smaller loop of radius $R_2$ is placed within the magnetic field of the larger loop.
The magnetic flux $\phi_2$ linked with the smaller loop is $\phi_2 = B \cdot A_2$, where $A_2 = \pi R_2^2$ is the area of the smaller loop.
Substituting the values, we get $\phi_2 = \left( \frac{\mu_0 I_1}{2 R_1} \right) (\pi R_2^2) = \left( \frac{\mu_0 \pi R_2^2}{2 R_1} \right) I_1$.
By definition, the mutual inductance $M$ is given by $\phi_2 = M I_1$, so $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.
Thus, $M$ is directly proportional to $\frac{R_2^2}{R_1}$.

(D) The magnetic field produced by the larger loop of radius $R_1$ at its center is $B = \frac{\mu_0 I}{2 R_1}$.
Assuming the smaller loop of radius $R_2$ is placed in this uniform magnetic field,the magnetic flux linked with the smaller loop is $\phi = B \cdot A_2$,where $A_2 = \pi R_2^2$ is the area of the smaller loop.
$\phi = \left( \frac{\mu_0 I}{2 R_1} \right) (\pi R_2^2) = \left( \frac{\mu_0 \pi R_2^2}{2 R_1} \right) I$.
Since $\phi = M I$,comparing the two expressions,we get $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.
Therefore,the mutual inductance $M$ is directly proportional to $\frac{R_2^2}{R_1}$.

(A) The magnetic field $B$ produced by a circular current-carrying loop is directed along its axis.
When an electron is projected along the axis of the circular conductor,its velocity vector $v$ is parallel to the magnetic field vector $B$.
Therefore,the angle $\theta$ between the velocity vector $v$ and the magnetic field $B$ is $0^{\circ}$.
The magnetic force $F$ experienced by a charged particle is given by $F = qvB \sin \theta$.
Substituting $\theta = 0^{\circ}$,we get $F = qvB \sin(0^{\circ}) = 0$.
Thus,the electron experiences no force.

(B) In a meter bridge experiment,the condition for the null point is given by the Wheatstone bridge principle: $\frac{P}{Q} = \frac{R_1}{R_2}$,where $R_1$ and $R_2$ are the resistances of the two segments of the wire.
Since $R = \rho \frac{L}{A}$,the ratio of the resistances of the two segments is $\frac{R_1}{R_2} = \frac{\rho (l/A)}{\rho ((100-l)/A)} = \frac{l}{100-l}$.
This ratio depends only on the lengths of the segments and is independent of the resistivity $\rho$ and the cross-sectional area $A$ of the wire.
Therefore,changing the cross-sectional area of the wire does not affect the balance condition.
The null point will remain at the same position,i.e.,$l \text{ cm}$ from the left end.

(B) In a meter bridge,the condition for the null point is given by the ratio: $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap and $Q$ is the resistance in the right gap.
Initially,$\frac{P}{Q} = \frac{l}{100-l}$.
When the resistances are doubled,they become $2P$ and $2Q$. When they are interchanged,the new left resistance becomes $2Q$ and the new right resistance becomes $2P$.
The new null point $l'$ satisfies: $\frac{2Q}{2P} = \frac{l'}{100-l'}$.
Simplifying the ratio: $\frac{Q}{P} = \frac{l'}{100-l'}$.
Since $\frac{P}{Q} = \frac{l}{100-l}$,then $\frac{Q}{P} = \frac{100-l}{l}$.
Equating the two: $\frac{l'}{100-l'} = \frac{100-l}{l}$.
By cross-multiplication: $l cdot l' = (100-l)(100-l')$.
$l cdot l' = 10000 - 100l' - 100l + l cdot l'$.
$100l' = 10000 - 100l$.
$l' = 100 - l$.
Therefore,the new position of the null point is $(100-l)$.

(C) For a meter bridge,the balancing condition is given by $\frac{X}{Y} = \frac{l}{100-l}$.
In the first case,$l = 20 \text{ cm}$.
$\frac{X}{Y} = \frac{20}{100-20} = \frac{20}{80} = \frac{1}{4}$.
Therefore,$Y = 4X$.
In the second case,the null point shifts to $l' = 40 \text{ cm}$. Since $X < Y$,we connect the $20 \text{ } \Omega$ resistor in series with $X$.
So,$X' = X + 20$ and $Y' = Y = 4X$.
The new balancing condition is $\frac{X+20}{4X} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.
Cross-multiplying gives $3(X + 20) = 2(4X)$.
$3X + 60 = 8X$.
$5X = 60$,which gives $X = 12 \text{ } \Omega$.
Since $Y = 4X = 48 \text{ } \Omega$,the smaller resistance is $X = 12 \text{ } \Omega$.

(A) Given: $\frac{R}{C_{V}} = 0.4$
$C_{V} = \frac{R}{0.4} = \frac{R}{2/5} = \frac{5R}{2}$
We know that for an ideal gas,$C_{P} = C_{V} + R$.
Substituting the value of $C_{V}$:
$C_{P} = \frac{5R}{2} + R = \frac{7R}{2}$
The adiabatic index $\gamma$ is given by $\gamma = \frac{C_{P}}{C_{V}}$.
$\gamma = \frac{7R/2}{5R/2} = \frac{7}{5} = 1.4$
For a gas,the degree of freedom $f$ is related to $\gamma$ by $\gamma = 1 + \frac{2}{f}$.
$1.4 = 1 + \frac{2}{f} \implies 0.4 = \frac{2}{f} \implies f = \frac{2}{0.4} = 5$.
$A$ gas with $f = 5$ degrees of freedom corresponds to rigid diatomic molecules.

(D) The pressure $P$ exerted by a gas on the walls of a container is defined as the force $F$ per unit area $A$,given by $P = \frac{F}{A}$.
Since the area $A$ of the container is constant,the force $F$ is directly proportional to the pressure $P$,i.e.,$F \propto P$.
From the ideal gas equation,$PV = nRT$,we can write $P = \frac{nRT}{V}$.
For a closed container,the volume $V$ is constant. Therefore,$P \propto T$.
Substituting this into the proportionality for force,we get $F \propto T$.
Comparing this with the given expression $F \propto T^{x}$,we find that $x = 1$.

(C) The root-mean-square (rms) velocity of gas molecules is given by the formula:
$v_{rms} = \sqrt{\frac{3kT}{m}}$
From this expression,we can see that $v_{rms} \propto \sqrt{T}$ for a given gas.
In an isothermal process,the temperature $(T)$ of the system remains constant.
Since the temperature does not change,the r.m.s. velocity of the gas molecules remains the same.

(B) Given: Area of square loop $A = 25 \, cm^2 = 25 \times 10^{-4} \, m^2$.
Side length $l = \sqrt{A} = 5 \, cm = 0.05 \, m$.
Resistance $R = 10 \, \Omega$, Time $t = 1 \, s$, Magnetic field $B = 40 \, T$.
The velocity $v$ required to pull the loop out in $1 \, s$ is $v = l/t = 0.05/1 = 0.05 \, m/s$.
The induced motional emf is $\varepsilon = Blv$.
The induced current is $I = \varepsilon/R = (Blv)/R$.
Substituting the values: $I = (40 \times 0.05 \times 0.05) / 10 = 0.01 \, A$.
The magnetic force acting on the loop is $F = BIl$.
Substituting the values: $F = 40 \times 0.01 \times 0.05 = 0.02 \, N$.
The work done is $W = F \times l = 0.02 \times 0.05 = 1 \times 10^{-3} \, J$.

(D) For a charged particle moving in a uniform magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$q v B = \frac{m v^2}{R}$
$\therefore R = \frac{m v}{q B}$
Given that the new speed $v' = \frac{v}{2}$ and the new magnetic field $B' = 2 B$.
The new radius $R'$ is given by:
$R' = \frac{m v'}{q B'} = \frac{m (v/2)}{q (2 B)}$
$R' = \frac{1}{4} \frac{m v}{q B}$
Since $R = \frac{m v}{q B}$,we get:
$R' = \frac{R}{4}$

(B) The magnetic field produced by a circular current-carrying loop along its axis is directed along the axis itself.
Since the electron is projected along the axis,its velocity vector $\vec{v}$ is parallel or antiparallel to the magnetic field vector $\vec{B}$.
The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$,which has a magnitude $F = qvB \sin \theta$.
Here,the angle $\theta$ between the velocity and the magnetic field is $0^{\circ}$ or $180^{\circ}$.
Since $\sin(0^{\circ}) = 0$ and $\sin(180^{\circ}) = 0$,the magnetic force $F = 0$.
Therefore,the electron experiences no force.

(C) An electromagnet requires a material that can be easily magnetized and demagnetized.
$1$. High retentivity allows the material to acquire a strong magnetic field when current flows through the coil.
$2$. Low coercivity ensures that the material loses its magnetism quickly when the current is switched off,allowing for efficient control.
Therefore,materials like soft iron are ideal for electromagnets because they possess high retentivity and low coercivity.

(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} MR^2$.
Using the equation of rotational motion,$\omega = \omega_0 + \alpha t$. Since the disc starts from rest,$\omega_0 = 0$,so $\alpha = \frac{\omega}{t}$.
The torque $\tau$ required is given by $\tau = I \alpha = (\frac{1}{2} MR^2) \times (\frac{\omega}{t}) = \frac{MR^2 \omega}{2t}$.
Since the force $F$ is applied tangentially at the rim,the torque is $\tau = F \times R$.
Equating the two expressions for torque: $F \times R = \frac{MR^2 \omega}{2t}$.
Therefore,the tangential force is $F = \frac{MR \omega}{2t}$.

(C) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$.
Given the angular frequency $\omega = \frac{2\pi}{T}$.
At position $x = \frac{a}{2}$,the speed is:
$v = \omega \sqrt{a^2 - (\frac{a}{2})^2}$
$v = \omega \sqrt{a^2 - \frac{a^2}{4}}$
$v = \omega \sqrt{\frac{3a^2}{4}}$
$v = \omega \cdot \frac{a\sqrt{3}}{2}$
Substituting $\omega = \frac{2\pi}{T}$:
$v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2}$
$v = \frac{\pi a \sqrt{3}}{T}$

(B) Given: Refractive index of glass $\mu_g = 1.5$,refractive index of air $\mu_a = 1$,refractive index of liquid $\mu_l = 1.25$,and radii of curvature $R_1 = 20 ~cm, R_2 = -20 ~cm$.
Using the Lens Maker's Formula: $P = \frac{1}{f} = (\frac{\mu_g}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
Power in air $(P_a)$: $P_a = (\frac{1.5}{1} - 1)(\frac{1}{20} - \frac{1}{-20}) = (0.5)(\frac{2}{20}) = 0.5 \times 0.1 = 0.05 ~cm^{-1}$.
Power in liquid $(P_l)$: $P_l = (\frac{1.5}{1.25} - 1)(\frac{1}{20} - \frac{1}{-20}) = (1.2 - 1)(0.1) = 0.2 \times 0.1 = 0.02 ~cm^{-1}$.
Ratio of power in liquid to power in air: $\frac{P_l}{P_a} = \frac{0.02}{0.05} = \frac{2}{5}$.

(B) The resolving power $R$ of a microscope is given by the formula: $R = \frac{2n \sin \alpha}{0.61 \lambda}$,where $n$ is the refractive index of the medium between the object and the objective lens,$\alpha$ is the semi-vertical angle,and $\lambda$ is the wavelength of light used.
From the formula,it is clear that $R \propto n$.
Therefore,by increasing the refractive index $n$ of the medium between the object and the objective lens,the resolving power of the microscope can be increased.

(A) The normal shift $x$ produced by a glass slab of thickness $t$ and refractive index $\mu$ is given by the formula:
$x = t \left( 1 - \frac{1}{\mu} \right)$
Rearranging the terms to solve for $t$:
$x = t \left( \frac{\mu - 1}{\mu} \right)$
$t = \frac{x \mu}{\mu - 1}$
Thus,the thickness of the glass slab is $\frac{\mu x}{\mu - 1}$.

(D) The kinetic energy of rotation $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.
Rearranging for angular momentum,we get $L = \sqrt{2KI}$.
Since the kinetic energies $K$ are equal for both bodies,we have $L \propto \sqrt{I}$.
Therefore,the ratio of the angular momenta of the two bodies is $\frac{L_1}{L_2} = \sqrt{\frac{I_1}{I_2}}$.
Given $I_1 = I$ and $I_2 = 2I$,we substitute these values:
$\frac{L_1}{L_2} = \sqrt{\frac{I}{2I}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1: \sqrt{2}$.

(A) The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega}{t}$.
The moment of inertia of a disc about its central axis is $I = \frac{1}{2} mR^2$.
The torque $\tau$ required is $\tau = I \alpha = (\frac{1}{2} mR^2) (\frac{\omega}{t}) = \frac{mR^2 \omega}{2t}$.
Since torque is also given by $\tau = F \cdot R$,where $F$ is the tangential force applied at the rim,
$F \cdot R = \frac{mR^2 \omega}{2t}$.
Solving for $F$,we get $F = \frac{mR \omega}{2t}$.

(B) According to the law of conservation of energy for incident radiation:
Coefficient of transmission $(t)$ + Coefficient of absorption $(a)$ + Coefficient of reflection $(r)$ $= 1$
Given: $a = 0.8$,$r = 0.1$
Therefore,$t = 1 - 0.8 - 0.1 = 0.1$
This means $10\%$ of the incident energy is transmitted through the surface.
Total incident energy in $5 ~min = 1000 ~J/min \times 5 ~min = 5000 ~J$
Transmitted energy $= 5000 ~J \times 0.1 = 500 ~J$

(C) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Therefore,$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
For a monoatomic ideal gas,the adiabatic index $\gamma = \frac{5}{3}$.
Thus,$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Since the gas is in a cylinder of constant cross-sectional area $A$,the volume $V$ is proportional to the length $L$ of the gas column $(V = AL)$.
Substituting $V_1 = AL_1$ and $V_2 = AL_2$,we get $\frac{V_1}{V_2} = \frac{L_1}{L_2}$.
Substituting these values into the temperature ratio equation:
$\frac{T_2}{T_1} = \left(\frac{L_1}{L_2}\right)^{2/3}$.

(B) The total incident energy per unit time is $Q_{in} = 1000 ~J/min$.
According to the law of conservation of energy,the sum of the coefficients of absorption $(a)$,reflection $(r)$,and transmission $(t)$ is equal to $1$:
$a + r + t = 1$
Given $a = 0.8$ and $r = 0.1$,we can find the coefficient of transmission $(t)$:
$0.8 + 0.1 + t = 1$
$0.9 + t = 1$
$t = 0.1$
Now,the rate of heat energy transmitted is $H = t \times Q_{in} = 0.1 \times 1000 = 100 ~J/min$.
The total heat energy transmitted in $5$ minutes is $E = H \times 5 = 100 \times 5 = 500 ~J$.

(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Given,initial volume $= V_i$,initial temperature $= T_i$,and final volume $V_f = \frac{V_i}{32}$.
Using the adiabatic relation: $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
Substituting the values: $T_i V_i^{\frac{7}{5}-1} = T_f \left(\frac{V_i}{32}\right)^{\frac{7}{5}-1}$.
$T_i V_i^{\frac{2}{5}} = T_f \left(\frac{V_i}{32}\right)^{\frac{2}{5}}$.
$T_i = T_f \left(\frac{1}{32}\right)^{\frac{2}{5}}$.
$T_f = T_i \times (32)^{\frac{2}{5}}$.
Since $32 = 2^5$,we have $(32)^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4$.
Therefore,$T_f = 4T_i$.
Comparing this with $T_f = xT_i$,we get $x = 4$.

(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Since the gas is monoatomic,the adiabatic index $\gamma = \frac{5}{3}$.
Therefore,$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
The volume of the gas column in a cylinder of cross-sectional area $A$ is $V = A \times L$.
Thus,$T_1 (A L_1)^{\gamma-1} = T_2 (A L_2)^{\gamma-1}$.
Rearranging for the ratio $\frac{T_1}{T_2}$,we get:
$\frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{\gamma-1} = \left(\frac{L_2}{L_1}\right)^{2/3}$.

(A) The position of the $n^{\text{th}}$ bright fringe is given by $y_{n} = \frac{n \lambda_{1} D}{d}$. For the $5^{\text{th}}$ bright band,$y_{5} = \frac{5 \lambda_{1} D}{d}$.
The position of the $m^{\text{th}}$ dark fringe is given by $y'_{m} = \frac{(2m - 1) \lambda_{2} D}{2d}$. For the $6^{\text{th}}$ dark band,$y'_{6} = \frac{(2 \times 6 - 1) \lambda_{2} D}{2d} = \frac{11 \lambda_{2} D}{2d}$.
Since the bands coincide,$y_{5} = y'_{6}$.
$\frac{5 \lambda_{1} D}{d} = \frac{11 \lambda_{2} D}{2d}$.
Simplifying,$5 \lambda_{1} = \frac{11 \lambda_{2}}{2}$.
Therefore,$\frac{\lambda_{1}}{\lambda_{2}} = \frac{11}{10}$,which implies $\frac{\lambda_{2}}{\lambda_{1}} = \frac{10}{11}$.

(D) Let the two slits be $S_1$ and $S_2$. The point $P$ on the screen is directly opposite to slit $S_1$.
The distance $S_1 P = D$ and the distance $S_2 P = \sqrt{D^2 + d^2}$.
Using the binomial expansion for $S_2 P$:
$S_2 P = D \left(1 + \frac{d^2}{D^2}\right)^{1/2} \approx D \left(1 + \frac{1}{2} \frac{d^2}{D^2}\right) = D + \frac{d^2}{2D}$.
The path difference $\Delta x$ between the two waves reaching point $P$ is:
$\Delta x = S_2 P - S_1 P = (D + \frac{d^2}{2D}) - D = \frac{d^2}{2D}$.
For a dark fringe,the path difference must be an odd multiple of $\frac{\lambda}{2}$. For the first dark fringe (minimum path difference),we set:
$\Delta x = \frac{\lambda}{2}$.
Equating the two expressions:
$\frac{d^2}{2D} = \frac{\lambda}{2}$.
Therefore,the wavelength of light is $\lambda = \frac{d^2}{D}$.

(C) The position of the $n^{\text{th}}$ bright band is given by $y_n = \frac{n \lambda_1 D}{d}$.
For the $5^{\text{th}}$ bright band $(n=5)$: $y_5 = \frac{5 \lambda_1 D}{d}$.
The position of the $n^{\text{th}}$ dark band is given by $y_n' = \frac{(2n-1) \lambda_2 D}{2d}$.
For the $6^{\text{th}}$ dark band $(n=6)$: $y_6' = \frac{(2 \times 6 - 1) \lambda_2 D}{2d} = \frac{11 \lambda_2 D}{2d}$.
Given that the bands coincide,$y_5 = y_6'$,so:
$\frac{5 \lambda_1 D}{d} = \frac{11 \lambda_2 D}{2d}$.
Canceling $D$ and $d$ from both sides:
$5 \lambda_1 = \frac{11 \lambda_2}{2}$.
Rearranging to find the ratio $\frac{\lambda_2}{\lambda_1}$:
$\frac{\lambda_2}{\lambda_1} = \frac{5 \times 2}{11} = \frac{10}{11}$.

(A) For a simple cubic unit cell, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $a = 2r$.
Therefore, $r = \frac{a}{2}$.
Given $a = 334.7 \ pm$.
$r = \frac{334.7 \ pm}{2} = 167.35 \ pm$.

(A) The $ccp$ (cubic close-packed) structure is also known as $fcc$ (face-centered cubic) structure.
Among the given options,$Cu$ (Copper) crystallizes in the $fcc$ $(ccp)$ structure.
$Zn$ (Zinc) and $Mg$ (Magnesium) crystallize in $hcp$ (hexagonal close-packed) structure.
$Po$ (Polonium) crystallizes in a simple cubic structure.

(B) The density formula for a unit cell is given by $\rho = \frac{M \cdot n}{a^3 \cdot N_A}$.
For an $fcc$ unit cell,the number of atoms per unit cell $(n)$ is $4$.
Given: $\rho = 21 \ g \ cm^{-3}$ and $a^3 \cdot N_A = 36 \ cm^3 \ mol^{-1}$.
Substituting the values: $21 = \frac{M \times 4}{36}$.
Solving for $M$: $M = \frac{21 \times 36}{4} = 21 \times 9 = 189.00 \ g \ mol^{-1}$.

(C) For an $fcc$ unit cell,the number of atoms per unit cell,$n = 4$.
The formula for density is $\rho = \frac{M \times n}{a^3 \cdot N_A}$.
Given $\rho = 2.7 \ g \ cm^{-3}$ and $a^3 \cdot N_A = 40 \ cm^3 \ mol^{-1}$.
Substituting the values: $2.7 = \frac{M \times 4}{40}$.
Solving for $M$: $M = \frac{2.7 \times 40}{4} = 27 \ g \ mol^{-1}$.

(B) For a $bcc$ unit cell, the relationship between radius $r$ and edge length $a$ is given by $r = \frac{\sqrt{3}}{4} a$.
Therefore, $a = \frac{4r}{\sqrt{3}}$.
Substituting the given value $r = 173 \ pm$:
$a = \frac{4 \times 173}{1.732} \approx 400 \ pm$.
Converting to centimeters:
$a = 400 \times 10^{-10} \ cm = 4.00 \times 10^{-8} \ cm$.

(D) For $fcc$ crystal structure, the relation between edge length $a$ and atomic radius $r$ is given by $4r = \sqrt{2}a$.
Therefore, $r = \frac{\sqrt{2}a}{4}$.
Substituting the given value $a = 393 \ pm$:
$r = \frac{1.414 \times 393}{4} = 138.93 \ pm$.

(B) Number of moles $= \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$
Total number of atoms $= 0.01 \times 6.022 \times 10^{23} = 6.022 \times 10^{21} \ \text{atoms}$
Number of atoms per unit cell $= 4$
Number of unit cells $= \frac{\text{Total number of atoms}}{\text{Atoms per unit cell}} = \frac{6.022 \times 10^{21}}{4} \approx 1.5 \times 10^{21} \ \text{unit cells}$

(C) For a $BCC$ unit cell, the number of atoms per unit cell is $n = 2$.
Density formula: $\rho = \frac{M \times n}{a^3 \times N_A}$
Rearranging for molar mass $M$: $M = \frac{\rho \times a^3 \times N_A}{n}$
Given: $\rho = 10 \ g \ cm^{-3}$, $a = 300 \ pm = 3 \times 10^{-8} \ cm$, $N_A = 6.022 \times 10^{23} \ mol^{-1}$, $n = 2$.
$M = \frac{10 \times (3 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2}$
$M = \frac{10 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{2}$
$M = \frac{162.594}{2} \approx 81.3 \ g \ mol^{-1}$.

(D) In a $bcc$ (body-centered cubic) unit cell,the atoms touch each other along the body diagonal.
The length of the body diagonal is given by $\sqrt{3} a$.
Since the body diagonal consists of two radii of the corner atoms and one full diameter of the central atom,we have $\sqrt{3} a = 4r$.
Rearranging this for the edge length $a$,we get $a = \frac{4r}{\sqrt{3}}$.

(C) For $fcc$ crystal structure, the relationship between radius $r$ and edge length $a$ is given by $r = \frac{a}{2\sqrt{2}} = \frac{\sqrt{2}}{4} a$.
Given $a = 405 \ pm$.
Substituting the value: $r = \frac{1.414 \times 405}{4} = \frac{572.67}{4} = 143.1675 \ pm \approx 143.2 \ pm$.

(A) For a simple cubic unit cell,the number of atoms per unit cell is $n = 1$.
The formula for density $(\rho)$ is given by $\rho = \frac{M \cdot n}{a^3 \cdot N_A}$.
Substituting the given values: $9.3 = \frac{M \cdot 1}{22.6}$.
Solving for molar mass $M$: $M = 9.3 \times 22.6 = 210.2 \ g \ mol^{-1}$.

(A) For a $bcc$ unit cell,the number of atoms per unit cell is $n = 2$.
The formula for density is $\rho = \frac{M \cdot n}{a^3 \cdot N_{A}}$.
Given $\rho = 7.8 \ g \ cm^{-3}$ and $a^3 \cdot N_{A} = 16.2 \ cm^3 \ mol^{-1}$.
Substituting the values: $7.8 = \frac{M \times 2}{16.2}$.
Solving for $M$: $M = \frac{7.8 \times 16.2}{2} = 63.18 \ g \ mol^{-1}$.

(A) Number of atoms in $0.3 \ mol = 0.3 \times N_A = 0.3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{23}$.
$I$. For $hcp$ structure,the number of octahedral voids is equal to the number of atoms.
Number of octahedral voids $= 1.8066 \times 10^{23}$.
$II$. For $hcp$ structure,the number of tetrahedral voids is equal to twice the number of atoms.
Number of tetrahedral voids $= 2 \times 1.8066 \times 10^{23} = 3.6132 \times 10^{23}$.

(B) In a simple cubic unit cell, the atoms touch each other along the edge of the cube.
Therefore, the relationship between the edge length $a$ and the atomic radius $r$ is given by $a = 2r$.
Given $r = 167.3 \ pm$.
Thus, $a = 2 \times 167.3 \ pm = 334.6 \ pm$.

(A) The density $\rho$ is given by the formula: $\rho = \frac{M \times Z}{a^3 \cdot N_A}$
For an $fcc$ unit cell,the number of atoms per unit cell,$Z = 4$.
Given: $M = 27 \ g \ mol^{-1}$ and $a^3 \cdot N_A = 38.5 \ cm^3 \ mol^{-1}$.
Substituting the values: $\rho = \frac{27 \ g \ mol^{-1} \times 4}{38.5 \ cm^3 \ mol^{-1}}$
$\rho = \frac{108}{38.5} \ g \ cm^{-3} \approx 2.8 \ g \ cm^{-3}$.

(A) For a $bcc$ unit cell, the relationship between the radius $r$ and the edge length $a$ is given by $4r = \sqrt{3}a$.
Substituting the given value of $a = 287 \ pm$:
$r = \frac{\sqrt{3}}{4} \times 287$
$r = \frac{1.732 \times 287}{4} = 124.27 \ pm$.

(B) $Zn$ (Zinc) crystallizes in a hexagonal close-packed $(hcp)$ structure.
$Cu$ (Copper) and $Ag$ (Silver) crystallize in a cubic close-packed ($ccp$ or $fcc$) structure.
$Po$ (Polonium) crystallizes in a simple cubic structure.

(B) For a $bcc$ unit cell, the relationship between the radius of the atom $(r)$ and the edge length $(a)$ is given by $4r = \sqrt{3}a$.
Substituting the given edge length $a = 450 \ pm$:
$r = \frac{\sqrt{3} \times 450}{4} = \frac{1.732 \times 450}{4} = 194.85 \ pm$.

(A) For an $fcc$ unit cell,the number of atoms per unit cell $(n)$ is $4$.
The density formula is given by $\rho = \frac{M \times n}{a^3 \times N_A}$.
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: $M = 180 \ g \ mol^{-1}$,$n = 4$,and $\rho \cdot N_A = 120 \times 10^{21} \ g \ cm^{-3} \ mol^{-1}$.
$a^3 = \frac{180 \times 4}{120 \times 10^{21}} = \frac{720}{120 \times 10^{21}} = 6.00 \times 10^{-21} \ cm^3$.

(A) The atoms of element $Y$ form an $hcp$ structure. Let the number of $Y$ atoms be $N$.
The number of tetrahedral voids generated is $2N$.
The atoms of element $X$ occupy $1/3$ of these tetrahedral voids.
Therefore,the number of $X$ atoms $= 2N \times 1/3 = 2/3 N$.
The ratio of $X$ atoms to $Y$ atoms is $(2/3 N) : N = 2/3 : 1 = 2 : 3$.
Thus,the formula of the compound is $X_2 Y_3$.

(D) The density of a unit cell is given by $\rho = \frac{M \cdot n}{a^3 \cdot N_A}$.
Here,$M = 56 \ g \ mol^{-1}$,$n = 2$ (for $bcc$ structure),and $\rho \cdot N_A = 4.8 \times 10^{24} \ g \ cm^{-3} \ mol^{-1}$.
The volume of the unit cell is $V = a^3 = \frac{M \cdot n}{\rho \cdot N_A}$.
Substituting the values: $V = \frac{56 \times 2}{4.8 \times 10^{24}} \ cm^3$.
$V = \frac{112}{4.8} \times 10^{-24} \ cm^3 = 23.33 \times 10^{-24} \ cm^3 = 2.33 \times 10^{-23} \ cm^3$.

(A) For a simple cubic unit cell,the number of atoms per unit cell $(n)$ is $1$.
Density $(\rho)$ is given by the formula: $\rho = \frac{n \cdot M}{a^3 \cdot N_{A}}$.
Substituting the given values: $\rho = \frac{1 \times 210}{21.5} \ g \ cm^{-3}$.
$\rho = 9.77 \ g \ cm^{-3}$.

(B) For a $bcc$ unit cell, the relationship between the radius $r$ and the edge length $a$ is given by $r = \frac{\sqrt{3}}{4} a$.
Therefore, $a = \frac{4r}{\sqrt{3}}$.
Substituting the value $r = 227 \ pm$:
$a = \frac{4 \times 227}{1.732} \approx 524.83 \ pm$.
Converting $pm$ to $cm$: $1 \ pm = 10^{-10} \ cm$.
$a = 524.83 \times 10^{-10} \ cm = 5.2483 \times 10^{-8} \ cm$.
Rounding to two decimal places, we get $5.24 \times 10^{-8} \ cm$.

(B) Concentration terms that involve volume are temperature-dependent because volume changes with temperature.
$Molarity$ $(M)$ is defined as the number of moles of solute per liter of solution $(mol/L)$.
Since volume is a component of the denominator,$Molarity$ changes with temperature.
In contrast,$Molality$,$Mole fraction$,and $Percent by mass$ are based on mass,which is independent of temperature.

(A) The solubility of gases in liquids depends on the nature of the gas and the solvent.
$O_2$ is a non-polar gas and exhibits very low solubility in water at room temperature.
In contrast,gases like $CO_2$,$NH_3$,and $HCl$ are highly soluble in water because they either react with water or are polar in nature.

(C) solution of ethyl alcohol in water is formed by dissolving a liquid solute (ethyl alcohol) in a liquid solvent (water). Therefore,it is classified as a $Liquid$ in $Liquid$ type of solution.

(A) The solubility of organic compounds in water depends on their ability to form hydrogen bonds with water molecules.
Alcohols $(R-OH)$ form stronger hydrogen bonds with water due to the high electronegativity of oxygen compared to nitrogen.
Amines $(R-NH_2)$ also form hydrogen bonds with water,but these are weaker than those formed by alcohols because nitrogen is less electronegative than oxygen.
Alkanes are non-polar and cannot form hydrogen bonds with water,making them practically insoluble.
Therefore,the decreasing order of solubility is: $\text{Alcohol} > \text{Amine} > \text{Alkane}$.

(D) Given: $n_2 = 1 \ mol$ (solute),$w_1 = 36 \ g$ (solvent),$P_1^0 = 32 \ mm \ Hg$.
Moles of water $(n_1)$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$.
According to Raoult's law for non-volatile solutes: $\frac{P_1^0 - P_1}{P_1^0} = \frac{n_2}{n_1 + n_2}$.
Substituting the values: $\frac{32 - P_1}{32} = \frac{1}{2 + 1} = \frac{1}{3}$.
$3(32 - P_1) = 32$.
$96 - 3P_1 = 32$.
$3P_1 = 96 - 32 = 64$.
$P_1 = \frac{64}{3} = 21.33 \ mm \ Hg$.

(D) For an ideal solution,the vapour pressure always lies between the vapour pressures of the pure components. Therefore,the statement that it is either higher or lower than the pure components is false.

(A) According to Henry's Law,the solubility $(S)$ of a gas is given by the formula: $S = K_{H} \times P$.
Given: $K_{H} = 0.16 \ mol \ dm^{-3} \ atm^{-1}$ and $P = 0.18 \ atm$.
Substituting the values: $S = 0.16 \times 0.18 = 0.0288 \ mol \ dm^{-3}$.
Rounding to two significant figures,we get $0.029 \ mol \ dm^{-3}$.

(A) According to Henry's law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid surface.
Mathematically,$P = K_H \cdot x$,where $P$ is the partial pressure,$x$ is the mole fraction (solubility),and $K_H$ is Henry's law constant.

(A) According to Henry's law,the solubility $(S)$ of a gas is given by the formula:
$S = K_{H} \times P$
Given:
$K_{H} = 0.159 \ mol \ dm^{-3} \ bar^{-1}$
$P = 0.346 \ bar$
Calculation:
$S = 0.159 \ mol \ dm^{-3} \ bar^{-1} \times 0.346 \ bar = 0.055 \ mol \ dm^{-3}$
Thus,the solubility of the gas is $0.055 \ mol \ dm^{-3}$.

(A) According to Henry's law,the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S = K_{H} \times P$.
Here,$S = 0.028 \ mol \ dm^{-3}$ and $P = 0.346 \ bar$.
Therefore,the Henry's law constant $(K_{H})$ is calculated as:
$K_{H} = \frac{S}{P} = \frac{0.028 \ mol \ dm^{-3}}{0.346 \ bar} \approx 0.081 \ mol \ dm^{-3} \ bar^{-1}$.

(C) The formula for molar mass of a solute is given by: $M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$
Given: $K_b = 2.5 \ K \ kg \ mol^{-1}$,$W_2 = 3.5 \ g$,$W_1 = 100 \ g$,and $\Delta T_b = 0.35 \ K$.
Substituting the values: $M_2 = \frac{2.5 \times 3.5 \times 1000}{0.35 \times 100}$
$M_2 = \frac{8750}{35} = 250 \ g \ mol^{-1}$.

(D) The osmotic pressure $\pi$ is given by the formula $\pi = M \times R \times T$,where $M$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Since $R$ and $T$ are constant for both cases,we have the relationship $\frac{\pi_1}{M_1} = \frac{\pi_2}{M_2}$.
Given $\pi_1 = 4.9 \ atm$,$M_1 = 0.2 \ M$,and $\pi_2 = 1.5 \ atm$.
Substituting the values: $\frac{4.9}{0.2} = \frac{1.5}{M_2}$.
$M_2 = \frac{1.5 \times 0.2}{4.9} = \frac{0.3}{4.9} \approx 0.0612 \ M$.
Rounding to the nearest option,the concentration is $0.06 \ M$.

(D) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality.
$m = \frac{W_2 \times 1000}{M_2 \times W_1} = \frac{1 \ g \times 1000}{60 \ g \ mol^{-1} \times 100 \ g} = \frac{1}{6} \ mol \ kg^{-1}$.
Given $\Delta T_{f} = 0.3 \ K$.
Using $\Delta T_{f} = K_{f} \times m$:
$0.3 \ K = K_{f} \times \frac{1}{6} \ mol \ kg^{-1}$.
$K_{f} = 0.3 \times 6 = 1.8 \ K \ kg \ mol^{-1}$.

(B) The formula for osmotic pressure is $\pi = \frac{W_2 RT}{M_2 V}$.
Rearranging for molar mass $M_2$,we get $M_2 = \frac{W_2 RT}{\pi V}$.
Substituting the given values: $W_2 = 8 \ g$,$R = 0.082 \ atm \ dm^3 \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$\pi = 0.6 \ atm$,and $V = 2 \ dm^3$.
$M_2 = \frac{8 \times 0.082 \times 300}{0.6 \times 2} \ g \ mol^{-1}$.
$M_2 = \frac{196.8}{1.2} \ g \ mol^{-1} = 164 \ g \ mol^{-1}$.

(B) Boiling point elevation is a colligative property given by $\Delta T_b = i \cdot K_b \cdot m$. Assuming complete dissociation,we compare the product $i \cdot m$ for each solution:
$(A)$ For $0.1 \text{ m } AlCl_3$: $i = 4$,$i \cdot m = 4 \times 0.1 = 0.4$
$(B)$ For $0.01 \text{ m } MgCl_2$: $i = 3$,$i \cdot m = 3 \times 0.01 = 0.03$
$(C)$ For $1 \text{ m } KCl$: $i = 2$,$i \cdot m = 2 \times 1 = 2.0$
$(D)$ For $0.5 \text{ m } NaCl$: $i = 2$,$i \cdot m = 2 \times 0.5 = 1.0$
The lowest value of $i \cdot m$ is $0.03$,hence $0.01 \text{ m } MgCl_2$ has the lowest boiling point elevation.

(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_2 \times 1000}{M_2 \times W_1}$,where $W_2$ is the mass of the solute,$M_2$ is the molar mass of the solute,and $W_1$ is the mass of the solvent in grams.
Substituting the given values: $W_2 = 3.2 \ g$,$M_2 = 128 \ g \ mol^{-1}$,$W_1 = 80 \ g$,and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.
$\Delta T_{f} = \frac{4.8 \ K \ kg \ mol^{-1} \times 3.2 \ g \times 1000 \ g \ kg^{-1}}{128 \ g \ mol^{-1} \times 80 \ g}$.
$\Delta T_{f} = \frac{15360}{10240} \ K = 1.5 \ K$.

(B) The formula for osmotic pressure is $\pi = iMRT = \frac{i \times W_2 \times R \times T}{M_2 \times V}$.
Given values are: $i = 2.47$,$W_2 = 1.7 \ g$,$R = 0.082 \ dm^3 \ atm \ mol^{-1} \ K^{-1}$,$T = 300 \ K$,$M_2 = 111 \ g \ mol^{-1}$,and $V = 1.25 \ dm^3$.
Substituting these values into the formula:
$\pi = \frac{2.47 \times 1.7 \times 0.082 \times 300}{111 \times 1.25} \ atm$.
$\pi = \frac{103.3638}{138.75} \ atm$.
$\pi = 0.744 \ atm$.

(D) The molality $m$ of the solution is calculated as follows:
Mass of glucose $(W_2)$ $= 36 \ g$.
Molar mass of glucose $(M_2)$ $= 180 \ g \ mol^{-1}$.
Volume of solution $= 1 \ dm^3 = 1 \ L$.
Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the mass of the solvent (water) $W_1$ is $1000 \ g = 1 \ kg$.
Molality $m = \frac{W_2}{M_2 \times W_1 (\text{in } kg)} = \frac{36}{180 \times 1} = 0.2 \ mol \ kg^{-1} = \frac{2}{10} \ mol \ kg^{-1}$.
The elevation in boiling point is given by $\Delta T_b = K_b \times m = K_b \times \frac{2}{10} = \frac{2 \ K_b}{10}$.
The boiling point of water is $100^{\circ} C$.
Therefore,the boiling point of the solution $T_b = 100 + \Delta T_b = (100 + \frac{2 \ K_b}{10})^{\circ} C$.

(A) The formula for osmotic pressure is $\pi = iMRT$.
Given:
$i = 1.83$
$M = 0.2 \ M$
$T = 0^{\circ} C = 273 \ K$
$R = 0.082 \ dm^3 \ atm \ mol^{-1} \ K^{-1}$
Substituting the values:
$\pi = 1.83 \times 0.2 \times 0.082 \times 273$
$\pi = 8.196 \ atm \approx 8.2 \ atm$.

(A) The formula for boiling point elevation is $\Delta T_b = K_b \times m$,where $m$ is the molality.
Molality $m = \frac{W_2 \times 1000}{M_2 \times W_1}$,where $W_2 = 1.5 \ g$,$M_2 = 150 \ g \ mol^{-1}$,and $W_1 = 30 \ g$.
Substituting these into the equation $\Delta T_b = K_b \times \frac{W_2 \times 1000}{M_2 \times W_1}$:
$0.65 = K_b \times \frac{1.5 \times 1000}{150 \times 30}$
$0.65 = K_b \times \frac{1500}{4500}$
$0.65 = K_b \times \frac{1}{3}$
$K_b = 0.65 \times 3 = 1.95 \ K \ kg \ mol^{-1}$.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
Given: $\Delta T_{b} = 1.89 \ K$ and $K_{b} = 3.15 \ K \ kg \ mol^{-1}$.
Substituting the values: $1.89 = 3.15 \times m$.
Therefore,$m = \frac{1.89}{3.15} = 0.6 \ mol \ kg^{-1}$ or $0.6 \ m$.

(C) The formula for osmotic pressure is $\pi = M R T = \frac{n_2 R T}{V}$.
Given:
$n_2 = 0.025 \ mol$
$V = 100 \ mL = 0.1 \ dm^3$
$T = 300 \ K$
$R = 0.082 \ atm \ dm^3 \ mol^{-1} \ K^{-1}$
Substituting the values:
$\pi = \frac{0.025 \ mol \times 0.082 \ atm \ dm^3 \ mol^{-1} \ K^{-1} \times 300 \ K}{0.1 \ dm^3}$
$\pi = \frac{0.615}{0.1} \ atm = 6.15 \ atm$.

(D) The boiling point elevation $\Delta T_b$ is given by $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor and $m$ is the molality. The value of $i \times m$ determines the magnitude of elevation.
$A) \ 0.1 \ m \ NaCl: i \times m = 2 \times 0.1 = 0.2 \ m$
$B) \ 0.2 \ m \ KNO_3: i \times m = 2 \times 0.2 = 0.4 \ m$
$C) \ 0.1 \ m \ Na_2SO_4: i \times m = 3 \times 0.1 = 0.3 \ m$
$D) \ 0.05 \ m \ CaCl_2: i \times m = 3 \times 0.05 = 0.15 \ m$
Since $0.05 \ m \ CaCl_2$ has the lowest value of $i \times m$ $(0.15)$,it exhibits the minimum boiling point elevation.

(C) The depression in freezing point is given by $\Delta T_{f} = T_{f}^0 - T_{f}$.
Given $T_{f}^0 = 0^{\circ} C$ and $T_{f} = -0.51^{\circ} C$,so $\Delta T_{f} = 0 - (-0.51) = 0.51 \ K$.
The formula for depression in freezing point is $\Delta T_{f} = i \times K_{f} \times m$.
Substituting the values: $0.51 = i \times 1.86 \times 0.15$.
Solving for $i$: $i = \frac{0.51}{1.86 \times 0.15} = \frac{0.51}{0.279} \approx 1.828$.
Rounding to two decimal places,the van't Hoff factor $i$ is $1.82$.

(A) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given that $1 \ \text{mole}$ of solute is dissolved in $1 \ kg$ of solvent,the molality $m = \frac{1 \ \text{mol}}{1 \ \text{kg}} = 1 \ \text{mol} \ kg^{-1}$.
Substituting the values into the equation: $x \ K = K_{b} \times 1 \ \text{mol} \ kg^{-1}$.
Therefore,$K_{b} = x \ K \ kg \ mol^{-1}$.

(A) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality.
Molality $m = \frac{W_2 \times 1000}{M_2 \times W_1}$,where $W_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent.
Substituting the values: $1.75 = \frac{3 \times 5.6 \times 1000}{M_2 \times 50}$.
Rearranging for $M_2$: $M_2 = \frac{3 \times 5.6 \times 1000}{1.75 \times 50}$.
$M_2 = \frac{16800}{87.5} = 192 \ g \ mol^{-1}$.

(A) The formula for depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$.
Given: $\Delta T_f = 0 - (-0.43) = 0.43 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.1 \ m$.
Substituting the values: $0.43 = i \times 1.86 \times 0.1$.
Solving for $i$: $i = \frac{0.43}{0.186} \approx 2.31$.
Rounding to the nearest given option,the value is $2.3$.