What is the freezing point depression constant of a solvent, 50g of which contain 1g non-volatile solute (molar mass…

Year	2026	2025	2024	Total
Questions	14	20	11	45

Q51 jee_main_2026_22_january_evening Mole Fraction and Concentration Terms

At

T(K)

,

100 g

of

98% H₂SO₄ (w/w)

aqueous solution is mixed with

100 g

of

49% H₂SO₄ (w/w)

aqueous solution. What is the mole fraction of

H₂SO₄

in the resultant solution? (Given: Atomic mass

H = 1 u

;

S = 32 u

;

O = 16 u

) (Assume that temperature after mixing remains constant)

A. $0.9$ $0.9$
B. $0.1$ $0.1$
C. $0.337$ $0.337$
D. $0.663$ $0.663$

Solution

Related Formula

Mass of solute = Total mass of solution \times Percentage (w/w)100

Mole fraction (xA) = (nA)/(nA + nB)

where, $nA$ $n_A$ = moles of solute ( $H₂SO₄$ $\text{H}_2\text{SO}_4$) $nB$ $n_B$ = moles of solvent ( $H₂O$ $\text{H}_2\text{O}$)

Core Logic

Step 1: Calculate total weight of $H₂SO₄$ $\text{H}_2\text{SO}_4$:

Weight = (100 \times (98)/(100)) + (100 \times (49)/(100)) = 98 + 49 = 147 g

Step 2: Calculate total weight of $H₂O$ $\text{H}_2\text{O}$:

Total weight of solution = 100 + 100 = 200 g

Weight of H₂O = 200 - 147 = 53 g

Step 3: Calculate moles and mole fraction:

nH₂SO₄ = (147)/(98) = 1.5

nH₂O = (53)/(18) \approx 2.944

xH₂SO₄ = (1.5)/(1.5 + 2.944) = 0.337

Pattern Recognition

Sees: Mixing two solution concentrations (w/w). Shortcut: Sum the component masses directly to find total solute mass and total solvent mass, then apply standard mole fraction formula.

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Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry

Q52 jee_main_2026_23_january_morning Non-Ideal Solutions and Raoult's Law

Which one of the following graphs accurately represents the plot of partial pressure of

CS₂

vs its mole fraction in a mixture of acetone and

CS₂

at constant temperature?

A. $(1)$ $(1)$
B. $(2)$ $(2)$
C. $(3)$ $(3)$
D. $(4)$ $(4)$

Solution

Related Formula

PCS₂ > PCS₂^° \cdot XCS₂

Core Logic

A mixture of carbon disulfide ( $CS₂$ $CS_2$) and acetone ( $CH₃-CO-CH₃$ $CH_3-CO-CH_3$) shows a positive deviation from Raoult's Law. This happens because the dipole-dipole interactions between acetone and $CS₂$ $CS_2$ are weaker than the attractive interactions between the pure molecules themselves.

Step 1: Graph Interpretation

Due to positive deviation, the actual partial pressure curve will bulge upwards (convex) relative to the ideal straight line (Raoult's Law). The first graph precisely depicts this upward curve starting from zero and reaching $P^°CS₂$ $P^\circ_{CS_2}$.

Pattern Recognition

Acetone + Carbon disulfide = Positive Deviation. Positive deviation means vapor pressure is higher than expected, so the curve sags upwards above the straight-line prediction.

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Class 12 Chemistry: Solutions

Q72 jee_main_2026_23_january_evening Raoult's Law

Two liquids A and B form an ideal solution. At

320 K

, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is

500 mm Hg

. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by

20 mm Hg

. Vapour pressure (in

mm Hg

) of B in the pure state is ____. (Nearest integer)

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

PTotal = PA° XA + PB° XB

Core Logic

Case 1: Moles of A ( $nA$ $n_A$) = 3, Moles of B ( $nB$ $n_B$) = 1. Total moles = $3 + 1 = 4$ $3 + 1 = 4$. Mole fraction of A ( $XA$ $X_A$) = $(3)/(4)$ $\frac{3}{4}$, Mole fraction of B ( $XB$ $X_B$) = $(1)/(4)$ $\frac{1}{4}$. Vapour pressure ( $Pₛ$ $P_{\text{s}}$) = $500 mm Hg$ $500 \text{ mm Hg}$. Using Raoult's law:

500 = PA°((3)/(4)) + PB°((1)/(4))

2000 = 3PA° + PB° --- (Equation 1)

Step 1: Setting up Case 2

Case 2: 1 mol of A is further added. So, new moles of A ( $nA$ $n_A^{\prime}$) = 4. Moles of B remains 1. Total moles = $4 + 1 = 5$ $4 + 1 = 5$. Mole fraction of A ( $XA$ $X_A^{\prime}$) = $(4)/(5)$ $\frac{4}{5}$, Mole fraction of B ( $XB$ $X_B^{\prime}$) = $(1)/(5)$ $\frac{1}{5}$. Vapour pressure increases by $20 mm Hg$ $20 \text{ mm Hg}$, so new $Pₛ = 500 + 20 = 520 mm Hg$ . Using Raoult's law:

520 = PA°((4)/(5)) + PB°((1)/(5))

2600 = 4PA° + PB° --- (Equation 2)

Step 2: Solving for Pure Vapour Pressures

Subtract Equation 1 from Equation 2:

(4PA° + PB°) - (3PA° + PB°) = 2600 - 2000

PA° = 600 mm Hg

Substitute $PA°$ $P_A^{\circ}$ back into Equation 1:

3(600) + PB° = 2000

1800 + PB° = 2000

PB° = 200 mm Hg

Pattern Recognition

A standard two-equation Raoult's law system. Cross-multiply out the denominators immediately to get simple linear equations ( $Ax + By = C$ $Ax + By = C$) which allow straightforward elimination.

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Class 12 Chemistry: Solutions

Q55 jee_main_2026_24_january_morning Osmotic Pressure

A solution is prepared by dissolving

0.3 g

of a non-volatile non-electrolyte solute 'A' of molar mass

60 g mol⁻¹

and

0.9 g

of a non-volatile non-electrolyte solute 'B' of molar mass

180 g mol⁻¹

in

100 mL

H₂O

at

27°C

. Osmotic pressure of the solution will be [Given:

R = 0.082 L atm K⁻¹ mol⁻¹

]

A. $1.23 atm$ $1.23 \text{ atm}$
B. $2.46 atm$ $2.46 \text{ atm}$
C. $0.82 atm$ $0.82 \text{ atm}$
D. $1.47 atm$ $1.47 \text{ atm}$

Solution

Related Formula

π = Ctotal RT

where, $Ctotal = nA + nBVsolution (L)$

Core Logic

Mass of solute 'A' = $0.3 g$ $0.3 \text{ g}$ Moles of solute 'A' = $0.3 g60 g/mol = (1)/(200) mol$

Mass of solute 'B' = $0.9 g$ $0.9 \text{ g}$ Moles of solute 'B' = $0.9 gm180 g/mol = (1)/(200) mol$

Total moles = $(1)/(200) + (1)/(200) = (2)/(200) mol$

Volume of solution = $100 mL = 0.1 L$ $100 \text{ mL} = 0.1 \text{ L}$

Total molarity ( $C$ $C$) of all solutes:

C = (2/200)/(100) \times 1000 = (1)/(10) M

Step 1: Calculate Osmotic Pressure

$π = C RT$ $\pi = C RT$

π = (1)/(10) \times 0.082 \times 300

π = 2.46 atm

Pattern Recognition

For non-electrolytes, $i=1$ $i=1$. Add the moles of all solutes present to find the total effective molarity. Use $T = 300 K$ $T = 300 \text{ K}$ for $27°C$ $27^{\circ}\text{C}$.

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Class 12 Chemistry: Solutions

Q62 jee_main_2026_24_january_morning Colligative Properties

'W' g of a non-volatile electrolyte solid solute of molar mass 'M'

g mol⁻¹

when dissolved in

100 mL

water, decreases vapour pressure of water from

640 mm Hg

to

600 mm Hg

. If aqueous solution of the electrolyte boils at

375 K

and

Kb

for water is

0.52 K kg mol⁻¹

, then the mole fraction of the electrolyte solute

(X₂)

in the solution can be expressed as (Given : density of water =

1 g/mL

and boiling point of water =

373 K

)

A. $(1.3)/(8) \times (W)/(M)$ $\frac{1.3}{8} \times \frac{W}{M}$
B. $(16)/(2.6) \times (W)/(M)$ $\frac{16}{2.6} \times \frac{W}{M}$
C. $(2.6)/(16) \times (M)/(W)$ $\frac{2.6}{16} \times \frac{M}{W}$
D. $(1.3)/(8) \times (M)/(W)$ $\frac{1.3}{8} \times \frac{M}{W}$

Solution

Related Formula

Δ PP° = i \cdot Xsolute

Δ Tb = i \cdot Kb \cdot m

Core Logic

From Relative Lowering of Vapour Pressure: $P° = 640 mm Hg$ $P^{\circ} = 640 \text{ mm Hg}$ $Pₛ = 600 mm Hg$ $P_s = 600 \text{ mm Hg}$ $Δ P = 40 mm Hg$ $\Delta P = 40 \text{ mm Hg}$

Moles of solute $n = (W)/(M)$ $n = \frac{W}{M}$ Mole fraction $Xsolute = Δ PP° \cdot (1)/(i)$ (for dilute solutions, or properly $i \cdot Xsolute = Δ PP°$ as given in the pdf approach) $Δ PP° = i \cdot Xsolute Xsolute = (40)/(640) \times (1)/(i)$

Now, from Boiling Point Elevation: $Δ Tb = 375 - 373 = 2 K$ $\Delta T_b = 375 - 373 = 2 \text{ K}$ $m = moles of solutemass of solvent in kg = (W/M)/(100/1000) = (W)/(M) \times 10$ $Δ Tb = i \times Kb \times m$ $\Delta T_b = i \times K_b \times m$ $2 = i \times 0.52 \times ( (W/M)/(100) \times 1000 )$ $i = (2)/(5.2) \times (M)/(W)$ $i = \frac{2}{5.2} \times \frac{M}{W}$

Step 1: Calculate Mole Fraction

Substitute the value of $i$ $i$ into the $Xsolute$ $X_{\text{solute}}$ equation:

Xsolute = (40)/(640) \times (1)/(i)

Xsolute = (1)/(16) \times (5.2)/(2) \times (W)/(M)

Xsolute = (2.6)/(16) \times (W)/(M) = (1.3)/(8) \times (W)/(M)

Pattern Recognition

When an unknown electrolyte is involved, isolate the van't Hoff factor ( $i$ $i$) from one colligative property equation and substitute it into the other to eliminate $i$ $i$ and solve for the target variable.

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Class 12 Chemistry: Solutions

What is the freezing point depression constant of a solvent, 50g$50\mathrm{g}$ of which contain 1g$1\mathrm{g}$ non-volatile solute (molar mass 256g mol⁻¹$256\mathrm{g\,mol}^{-1}$ ) and the decrease in freezing point is 0.40K$0.40\mathrm{K}$ ?

Solution & Explanation

Related Formula

Step 1: Compute Molality

Step 2: Calculate Kf$K_f$

Pattern Recognition

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More Solutions Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

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Solution

Related Formula

Core Logic

Step 1: Graph Interpretation

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Setting up Case 2

Step 2: Solving for Pure Vapour Pressures

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Calculate Osmotic Pressure

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Mole Fraction

Pattern Recognition

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What is the freezing point depression constant of a solvent, $50g$ $50\mathrm{g}$ of which contain $1g$ $1\mathrm{g}$ non-volatile solute (molar mass $256g mol⁻¹$ $256\mathrm{g\,mol}^{-1}$ ) and the decrease in freezing point is $0.40K$ $0.40\mathrm{K}$ ?

Step 2: Calculate $Kf$ $K_f$