MHT CET 2024 Chemistry Question Paper with Answer and Solution

(C) The osmotic pressure formula is $\Pi = CRT$,where $C = \frac{n}{V} = \frac{W_2}{M_2 \times V}$.
Substituting the values into the formula: $\Pi = \frac{W_2 \times R \times T}{M_2 \times V}$.
Rearranging to solve for mass $(W_2)$: $W_2 = \frac{\Pi \times M_2 \times V}{R \times T}$.
$W_2 = \frac{0.245 \ atm \times 58 \ g \ mol^{-1} \times 2.5 \ dm^3}{0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1} \times 300 \ K}$.
$W_2 = \frac{35.525}{24.63} \approx 1.44 \ g$.

(B) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} \times m$.
First,calculate the molality $(m)$: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.
Now,calculate the elevation in boiling point: $\Delta T_{b} = 0.513 \ ^{\circ}C \ kg \ mol^{-1} \times 0.5 \ mol \ kg^{-1} = 0.2565 \ ^{\circ}C$.
The boiling point of the solution $(T)$ is $T_{b} + \Delta T_{b} = 100 \ ^{\circ}C + 0.2565 \ ^{\circ}C = 100.2565 \ ^{\circ}C$,which is approximately $100.256 \ ^{\circ}C$.

(C) The elevation in boiling point $(\Delta T_b)$ is a colligative property,which depends on the van't Hoff factor $(i)$ and the molality $(m)$ of the solution: $\Delta T_b = i \times K_b \times m$.
Upon complete dissociation,the number of particles produced per formula unit is:
$(A)$ $0.1 \ m \ KCl \rightarrow i = 2$,particles = $0.1 \times 2 = 0.2 \ m$
$(B)$ $0.05 \ m \ NaCl \rightarrow i = 2$,particles = $0.05 \times 2 = 0.1 \ m$
$(C)$ $0.1 \ m \ BaCl_2 \rightarrow i = 3$,particles = $0.1 \times 3 = 0.3 \ m$
$(D)$ $0.1 \ m \ MgSO_4 \rightarrow i = 2$,particles = $0.1 \times 2 = 0.2 \ m$
Since the $0.1 \ m \ BaCl_2$ solution has the highest concentration of particles $(0.3 \ m)$,it exhibits the maximum elevation in boiling point.

(B) The formula for molar mass of a solute is given by: $M_2 = \frac{K_{f} \times W_2 \times 1000}{\Delta T_{f} \times W_1}$.
Given values are: $W_2 = 1 \ g$,$W_1 = 100 \ g$,$\Delta T_{f} = 0.2 \ K$,and $K_{f} = 1.2 \ K \ kg \ mol^{-1}$.
Substituting these values into the formula:
$M_2 = \frac{1.2 \times 1 \times 1000}{0.2 \times 100} = \frac{1200}{20} = 60 \ g \ mol^{-1}$.

(B) The formula for the depression in freezing point is $\Delta T_{f} = i \times m \times K_{f}$.
Given:
van't Hoff factor $(i)$ = $1.1$
Molality $(m)$ = $0.01 \ m$
Cryoscopic constant $(K_{f})$ = $1.86 \ K \ kg \ mol^{-1}$
Substituting the values:
$\Delta T_{f} = 1.1 \times 0.01 \times 1.86$
$\Delta T_{f} = 0.02046 \ K \approx 0.020 \ K$.

(D) The formula for the depression in freezing point is $\Delta T_f = K_f \times m$,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 in grams.
Substituting the values: $\Delta T_f = 3 \ K$,$W_2 = 2.5 \ g$,$M_2 = 117 \ g \ mol^{-1}$,$W_1 = 35 \ g$.
$K_f = \frac{\Delta T_f \times M_2 \times W_1}{1000 \times W_2}$
$K_f = \frac{3 \times 117 \times 35}{1000 \times 2.5} \ K \ kg \ mol^{-1}$
$K_f = \frac{12285}{2500} \ K \ kg \ mol^{-1} = 4.91 \ K \ kg \ mol^{-1}$.

(A) Osmotic pressure is a colligative property that depends on the number of particles in the solution. The solution having a higher concentration of particles will have a larger osmotic pressure.
Osmotic pressure $(\pi)$ $\propto i \times C$,where $i$ is the van't Hoff factor and $C$ is the molarity.

Solution	Concentration of particles $(i \times m)$
$(a) \ 0.5 \ m \ Li_2SO_4$	$3 \times 0.5 = 1.5 \ m$
$(b) \ 0.5 \ m \ KCl$	$2 \times 0.5 = 1.0 \ m$
$(c) \ 0.5 \ m \ Al_2(SO_4)_3$	$5 \times 0.5 = 2.5 \ m$
$(d) \ 0.1 \ m \ BaCl_2$	$3 \times 0.1 = 0.3 \ m$

Comparing the values: $0.3 < 1.0 < 1.5 < 2.5$.
Therefore,the correct order of increasing osmotic pressure is $d < b < a < c$.

(D) The osmotic pressure formula is given by $\pi V = nRT = \frac{W_2}{M_2} RT$.
Rearranging for molar mass $M_2$: $M_2 = \frac{W_2 RT}{\pi V}$.
Given: $W_2 = 1 \ g$,$V = 0.3 \ dm^3$,$\pi = 0.2 \ atm$,$T = 300 \ K$,$R = 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $M_2 = \frac{1 \ g \times 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1} \times 300 \ K}{0.2 \ atm \times 0.3 \ dm^3}$.
$M_2 = \frac{24.6}{0.06} \ g \ mol^{-1} = 410 \ g \ mol^{-1}$.

(B) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
Given: $\Delta T_{b} = 0.5 \ K$ and $K_{b} = 2.40 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.5 \ K = 2.40 \ K \ kg \ mol^{-1} \times m$.
Therefore,$m = \frac{0.5}{2.40} \approx 0.2083 \ mol \ kg^{-1}$.
Rounding to two decimal places,we get $m = 0.21 \ mol \ kg^{-1}$.

(A) Given:
- For $0.1 \ molal$ solution,$T_b = 100.16^{\circ} C$.
- Boiling point of pure water is $100^{\circ} C$.
- Boiling point elevation formula: $\Delta T_b = K_b \cdot m$ (since glucose is a non-electrolyte,$i = 1$).
Step $1$: Calculate elevation for $0.1 \ molal$ solution:
$\Delta T_b = 100.16 - 100 = 0.16^{\circ} C$.
Step $2$: Use the proportionality $\Delta T_b \propto m$ to find elevation for $0.5 \ molal$ solution:
$\Delta T_b^{\prime} = \Delta T_b \times \frac{m^{\prime}}{m} = 0.16 \times \frac{0.5}{0.1} = 0.16 \times 5 = 0.8^{\circ} C$.
Step $3$: Calculate new boiling point:
$T_b^{\prime} = 100 + 0.8 = 100.80^{\circ} C$.
Thus,the correct option is $A$.

(D) The depression in freezing point is given by the formula: $\Delta T_{f} = i \times m \times K_{f}$
Given: $\Delta T_{f} = 0 - (-0.680^{\circ} C) = 0.680 \ K$,molality $m = 0.2 \ mol \ kg^{-1}$,and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.680 = i \times 0.2 \times 1.86$
$i = \frac{0.680}{0.2 \times 1.86} = \frac{0.680}{0.372} \approx 1.8279$
Rounding to two decimal places,we get $i \approx 1.83$.

(B) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \cdot m$,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 solute,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent.
Substituting the values: $\Delta T_{b} = 0.01 \ K$,$K_{b} = 0.50 \ K \ kg \ mol^{-1}$,$W_2 = 0.35 \ g$,and $W_1 = 100 \ g$.
$0.01 = 0.50 \times \frac{0.35 \times 1000}{M_2 \times 100}$.
$M_2 = \frac{0.50 \times 0.35 \times 1000}{0.01 \times 100} = \frac{175}{1} = 175 \ g \ mol^{-1}$.

(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 $(x_2)$.
The formula is: $\frac{P_1^0 - P_1}{P_1^0} = x_2$
Given: $P_1^0 = 550 \ mm \ Hg$ and $P_1 = 510 \ mm \ Hg$.
Substituting the values: $x_2 = \frac{550 - 510}{550} = \frac{40}{550} \approx 0.0727$.
Therefore,the mole fraction of the solute is approximately $0.072$.

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i \cdot K_{f} \cdot m$.
Given: $\Delta T_{f} = 0.660 \ K$,$m = 0.2 \ m$,and $K_{f} = 1.84 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.660 = i \times 1.84 \times 0.2$.
$i = \frac{0.660}{1.84 \times 0.2} = \frac{0.660}{0.368} \approx 1.793$.
Thus,the van't Hoff factor $(i)$ is approximately $1.79$.

(B) According to Raoult's law,the lowering of vapour pressure $(\Delta P)$ is directly proportional to the mole fraction of the solute $(X_{solute})$: $\Delta P = P^o \cdot X_{solute}$.
Given:
Case $1$: $\Delta P_1 = 10 \text{ mm Hg}$,$X_1 = 0.2$
Case $2$: $\Delta P_2 = 20 \text{ mm Hg}$,$X_2 = ?$
Since $\frac{\Delta P_1}{\Delta P_2} = \frac{X_1}{X_2}$,we have:
$\frac{10}{20} = \frac{0.2}{X_2}$
$X_2 = \frac{0.2 \times 20}{10} = 0.4$.

(D) For sucrose solution,$(\Delta T_f)_1 = K_f m_1 \dots (I)$.
For $CaCl_2$ solution,$(\Delta T_f)_2 = i K_f m_2 \dots (II)$.
Since $m_1 = m_2 = 1.25 \ m$,we have $\frac{(\Delta T_f)_2}{(\Delta T_f)_1} = \frac{i K_f m_2}{K_f m_1} = i$.
For $CaCl_2$,the van't Hoff factor $i = 3$ (as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$).
Therefore,$(\Delta T_f)_2 = i \times (\Delta T_f)_1 = 3 \times x \ K = 3 x \ K$.

(D) Given: Moles of solute $(n_2)$ = $0.1 \text{ mol}$, Mass of water $(w_1)$ = $16.2 \text{ g}$, Molar mass of water $(M_1)$ = $18 \text{ g mol}^{-1}$, $P_1^0 = 24 \text{ mmHg}$.
Moles of solvent $(n_1)$ = $\frac{w_1}{M_1} = \frac{16.2}{18} = 0.9 \text{ mol}$.
Using the formula for relative lowering of vapour pressure:
$\frac{P_1^0 - P_1}{P_1^0} = \frac{n_2}{n_1 + n_2}$
$\frac{24 - P_1}{24} = \frac{0.1}{0.9 + 0.1} = \frac{0.1}{1.0} = 0.1$
$24 - P_1 = 24 \times 0.1 = 2.4$
$P_1 = 24 - 2.4 = 21.6 \text{ mmHg}$.
Note: If using the approximation $\frac{P_1^0 - P_1}{P_1^0} = \frac{n_2}{n_1}$, then $\frac{24 - P_1}{24} = \frac{0.1}{0.9} = 0.111$, which gives $P_1 = 24 - 2.67 = 21.33 \text{ mmHg} \approx 21.3 \text{ mmHg}$. Given the options, the approximation is intended.

(C) The formula for osmotic pressure is $\pi = \frac{W_2 RT}{M_2 V}$.
Rearranging for molar mass: $M_2 = \frac{W_2 RT}{\pi V}$.
Given: $W_2 = 400 \ mg = 0.4 \ g$,$T = 300 \ K$,$V = 300 \ mL = 0.3 \ L$,$\pi = 0.2 \ atm$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $M_2 = \frac{0.4 \ g \times 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K}{0.2 \ atm \times 0.3 \ L}$.
$M_2 = \frac{9.852}{0.06} = 164.2 \ g \ mol^{-1} \approx 164 \ g \ mol^{-1}$.

(C) The relative lowering of vapour pressure is given by the formula: $\frac{\Delta P}{P_1^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W_2 \times M_1}{M_2 \times W_1}$.
Given:
Mass of solute $(W_2)$ $= 46 \ g$
Molar mass of solute $(M_2)$ $= 46 \ g \ mol^{-1}$
Mass of solvent $(W_1)$ $= 162 \ g$
Molar mass of water $(M_1)$ $= 18 \ g \ mol^{-1}$
Substituting the values:
$\frac{\Delta P}{P_1^0} = \frac{46 \times 18}{46 \times 162} = \frac{18}{162} = \frac{1}{9} \approx 0.111$.

(D) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{0} - T_{f} = 0 - (-0.5 \ K) = 0.5 \ K$.
Using the formula for depression in freezing point with the van't Hoff factor: $\Delta T_{f} = i \times K_{f} \times m$.
Assuming molarity $(M)$ is approximately equal to molality $(m)$ for dilute aqueous solutions,$m \approx 0.15 \ mol \ kg^{-1}$.
Rearranging for $i$: $i = \frac{\Delta T_{f}}{K_{f} \times m} = \frac{0.5 \ K}{1.86 \ K \ kg \ mol^{-1} \times 0.15 \ mol \ kg^{-1}}$.
Calculating the value: $i = \frac{0.5}{0.279} \approx 1.79$.

(B) Osmotic pressure is a colligative property,not osmosis itself. Thus,statement $A$ is incorrect.The vapour pressure of a solution containing a non-volatile solute is lower than that of the pure solvent due to the decrease in the number of solvent molecules at the surface. Thus,statement $B$ is correct.Osmotic pressure is a colligative property dependent on the number of particles. $NaCl$ dissociates into $2$ ions ($Na^+$ and $Cl^-$),while sucrose does not dissociate. Therefore,the osmotic pressure of $0.1 \ M$ $NaCl$ is higher than $0.1 \ M$ sucrose. Thus,statement $C$ is incorrect.The boiling point of a solution containing a non-volatile solute is higher than that of the pure solvent (elevation in boiling point). Thus,statement $D$ is incorrect.

(D) The formula for depression in freezing point is $\Delta T_f = K_f \times m$.
Given: $\Delta T_f = 0.18 \ K$ and $K_f = 1.6 \ K \ kg \ mol^{-1}$.
Substituting the values in the formula:
$m = \frac{\Delta T_f}{K_f} = \frac{0.18 \ K}{1.6 \ K \ kg \ mol^{-1}} = 0.1125 \ m$.
Rounding to three decimal places,we get $m = 0.113 \ m$.

(C) The relative lowering of vapour pressure is given by the formula: $\frac{P^0 - P}{P^0}$
Given:
$P^0 = 32 \ mm \ Hg$ (vapour pressure of pure solvent)
$P = 30 \ mm \ Hg$ (vapour pressure of solution)
Relative lowering of vapour pressure $= \frac{32 - 30}{32} = \frac{2}{32} = 0.0625$

(A) The formula for osmotic pressure is $\pi = CRT = \frac{W_2}{M_2 V} RT$.
Rearranging for molar mass $M_2$: $M_2 = \frac{W_2 RT}{\pi V}$.
Substituting the given values: $M_2 = \frac{4 \ g \times 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1} \times 300 \ K}{2 \ atm \times 1 \ dm^3}$.
$M_2 = \frac{98.4}{2} \ g \ mol^{-1} = 49.2 \ g \ mol^{-1}$.

(A) The electrode reaction for the evolution of chlorine gas is: $2 Cl^{-} \rightarrow Cl_2 + 2 e^{-}$.
According to Faraday's law,the number of moles of electrons passed is $n(e^{-}) = \frac{I \times t}{F} = \frac{1 \times 965}{96500} = 0.01 \ mol$.
From the stoichiometry of the reaction,$2 \ mol$ of $e^{-}$ produce $1 \ mol$ of $Cl_2$.
Therefore,moles of $Cl_2$ produced $= \frac{0.01}{2} = 0.005 \ mol$.
At $STP$,the volume of $1 \ mol$ of gas is $22.4 \ L$.
Volume of $Cl_2 = 0.005 \ mol \times 22.4 \ L \ mol^{-1} = 0.112 \ L$.

(A) Nitrogen has two naturally occurring stable isotopes,namely,$^{14}N$ and $^{15}N$.

(B) Isotones are atoms of different elements that have the same number of neutrons in their nuclei.
To find the number of neutrons,we use the formula: $n = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For ${ }_{6}^{12}C$: $n = 12 - 6 = 6$.
For ${ }_{5}^{11}B$: $n = 11 - 5 = 6$.
Since both have $6$ neutrons,they are isotones.

(A) $UV-visible$ spectroscopy is commonly used for the preliminary confirmation of the synthesis of nanoparticles,especially metal nanoparticles,due to the phenomenon of Surface Plasmon Resonance $(SPR)$.
$SPR$ causes a characteristic absorption peak in the $UV-visible$ region,which serves as a quick and simple indicator of nanoparticle formation.

(B) The atomic number $27$ corresponds to Cobalt $(Co)$.
The electronic configuration of neutral $Co$ $(Z=27)$ is $[Ar] \ 3d^7 \ 4s^2$.
When $Co$ is in the $+2$ oxidation state,it loses two electrons from the $4s$ orbital.
The resulting electronic configuration is $[Ar] \ 3d^7$.
Therefore,the number of electrons in the $d$-subshell is $7$.

(D) Absorption is a bulk phenomenon where the substance is uniformly distributed throughout the body of the material. Unlike adsorption,it is not a surface phenomenon and therefore does not depend on the surface area of the absorbent.

(A) According to the Freundlich adsorption isotherm equation: $\log \frac{x}{m} = \log K + \frac{1}{n} \log C$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log C$,and $c = \log K$.
The slope of the straight line is $\frac{1}{n}$.

(D) Heterogeneous catalysis is a process where the catalyst exists in a different physical phase than the reactants.
In the hydrogenation of vegetable oil,the reactant (vegetable oil) is in the liquid phase,while the catalyst $(Ni_{(s)})$ is in the solid phase.
Since the phases are different,this is an example of heterogeneous catalysis.
In options $A$,$B$,and $C$,the catalyst and the reactants are in the same phase (homogeneous catalysis).

(A) Cheese is a type of colloid known as a gel.
In a gel,the dispersed phase is a liquid and the dispersion medium is a solid.
According to the provided table,row $1$ represents a system where the dispersed phase is liquid and the dispersion medium is solid.
Therefore,the correct option is $1$.

(C) Multimolecular colloids are formed by the aggregation of a large number of atoms or smaller molecules with diameters less than $1 \ nm$.
Sulfur sol,which consists of a large number of $S_8$ molecules,is a classic example of a multimolecular colloid.
Soap and synthetic polymers like Polythene and Nylon are examples of associated colloids and macromolecular colloids,respectively.

(D) In macromolecular colloids,the molecules of the dispersed phase are sufficiently large in size (macro) to be of colloidal dimensions.
Examples of macromolecular colloids include starch,cellulose,proteins,polythene,nylon,and plastics.
Soap is an example of an associated colloid (micelle),not a macromolecular colloid.

(D) In lyophilic colloids,the particles of the dispersed phase have a great affinity for the dispersion medium.
They are reversible and self-stabilized colloids.
Addition of a large amount of electrolytes is required to cause precipitation or coagulation of lyophilic sols,whereas lyophobic colloids are easily coagulated by small amounts of electrolytes.

(D) According to the $Schulze-Hardy$ rule,the coagulating power of an ion is directly proportional to its valency (charge) for the precipitation of a sol.
For a positive sol,the coagulating power depends on the magnitude of the negative charge of the anion.
The order of coagulating power is $\left[Fe(CN)_6\right]^{4-} > PO_4^{3-} > SO_4^{2-} > Cl^{-}$.
Therefore,the anion with the lowest valency,$Cl^{-}$,has the lowest coagulating power.

(C) Milk is an emulsion,which is a type of colloid where a $liquid$ is dispersed in another $liquid$.
Specifically,it consists of $liquid$ butterfat globules dispersed within a water-based solution.

(C) Soap lather is a type of colloidal system known as $Foam$.
In this system,a gas is dispersed in a liquid medium.

(A) Fog is a type of colloid known as an aerosol where a liquid is dispersed in a gas.
Therefore,the dispersed phase is liquid and the dispersion medium is gas.
Thus,option $(A)$ is the correct answer.

(A) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to the magnitude of its charge.
For the precipitation of a positive sol,the anion with the highest negative charge will have the maximum coagulating power.
Comparing the charges: $\left[Fe(CN)_6\right]^{4-}$ has a charge of $-4$,$PO_4^{3-}$ has $-3$,$SO_4^{2-}$ has $-2$,and $Cl^{-}$ has $-1$.
Therefore,$\left[Fe(CN)_6\right]^{4-}$ has the maximum coagulating power.

(D) The classification of colloids based on the physical state of the dispersed phase and dispersion medium is as follows:
$1$. Foam rubber is a solid sol (gas in solid).
$2$. Froth is a foam (gas in liquid).
$3$. Gelatin is a gel (liquid in solid).
$4$. Hair cream is an emulsion (liquid in liquid).
Therefore,the correct option is $D$.