Assume a living cell with 0.9% (omega/omega) of glucose solution (aqueous). This cell is immersed in another solution…

Year	2026	2025	2024	Total
Questions	14	20	11	45

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Mass percentage from mole fraction calculation:

% w/w = (x₁ \cdot M₁)/(x₁ \cdot M₁ + x₂ \cdot M₂) \times 100

Core Logic

Inside the living cell, glucose concentration is $0.9% w/w$ $0.9\%\text{ w/w}$.

The surrounding solution has equal mole fractions of glucose and water ( $xglucose = 0.5$ $x_{\text{glucose}} = 0.5$, $xwater = 0.5$ $x_{\text{water}} = 0.5$).

Let's calculate the mass percentage of the outer solution:

Mass of glucose component = $0.5 \times 180 = 90 g$ $0.5 \times 180 = 90\mathrm{\ g}$
Mass of water component = $0.5 \times 18 = 9 g$ $0.5 \times 18 = 9\mathrm{\ g}$
Total solution mass = $90 + 9 = 99 g$ $90 + 9 = 99\mathrm{\ g}$

Step 1: Concentration Determination and Osmosis Profile

Outer mass percentage:

% w/w = (90)/(99) \times 100 \approx 90.9%

Because the external environment is highly concentrated (hypertonic) compared to the inner cell ( $0.9%$ $0.9\%$), water flows out of the cell via exosmosis, causing the cell to shrink.

Note: Because the reasoning establishes shrinkage but the quantitative figures in the options are highly mismatched, this question is officially designated as a Bonus question.

Pattern Recognition

An equal mole fraction solution of a high-molar-mass solute (glucose, $180 g/mol$ $180\mathrm{\ g/mol}$) and a low-molar-mass solvent (water, $18 g/mol$ $18\mathrm{\ g/mol}$) is always extremely concentrated by mass. Placing a standard living cell into such a hypertonic solution inevitably causes fluid loss and cellular shrinkage.

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Class 12 Chemistry: Solutions

More Solutions Previous-Year Questions — Page 8

Q jee_main_2024_29_january_evening Interconversion of Concentration Terms

Molality of

0.8 M

H₂SO₄

solution (density

1.06 g cm⁻³

) is **815**

\times 10⁻³ m

Numerical Answer. Answer: 815 to 815

Solution

Related Formula

m = (M \times 1000)/((1000 \times d) - (M \times MB))

where,

$M$ $M$ = Molarity = $0.8 M$ $0.8\text{ M}$
$d$ $d$ = Density = $1.06 g/cm³$ $1.06\text{ g/cm}^3$
$MB$ $M_B$ = Molar mass of solute ( $H₂SO₄$ $\mathrm{H}_2\mathrm{SO}_4$) = $98 g/mol$ $98\text{ g/mol}$

Core Logic

Substituting the given values into the equation:

m = (0.8 \times 1000)/((1000 \times 1.06) - (0.8 \times 98))

Calculating the denominator parameters:

Denominator = 1060 - 78.4 = 981.6 g

Step 1: Final Resolution

Solving for molality:

m = (800)/(981.6) \approx 0.815 m = 815 \times 10⁻³ m

Thus, the integer factor value is 815.

Pattern Recognition

Ensure you explicitly subtract the mass of the solute from the total mass of the solution to correctly isolate the mass of the solvent needed for molality calculations.

Chapter Mix

Class 12 Chemistry: Solutions

Q67 jee_main_2024_27_jan_morning Vapour Pressure and Deviations from Raoult's Law

A solution of two miscible liquids showing negative deviation from Raoult's law will have:

A. increased vapour pressure, increased boiling point
B. increased vapour pressure, decreased boiling point
C. decreased vapour pressure, decreased boiling point
D. decreased vapour pressure, increased boiling point

Solution

Core Logic

A system demonstrating a negative deviation from Raoult's law implies tighter molecular attractions between components ( $A-B$ $A-B$ interactions are stronger than $A-A$ $A-A$ or $B-B$ $B-B$). This decreases the aggregate escaping tendency, yielding a decreased total vapour pressure. Consequently, a higher thermal energy threshold is required to reach the boiling threshold, causing an increased boiling point.

Pattern Recognition

Negative deviation $arrow$ $\rightarrow$ Vapour Pressure drops $arrow$ $\rightarrow$ Boiling Point rises inversely.

Chapter Mix

Class 12 Chemistry: Solutions

Q85 jee_main_2024_29_jan_morning Concentration Terms

A solution of

H₂SO₄

31.4%

H₂SO₄

by mass and has a density of

1.25g / mL

. The molarity of the

H₂SO₄

solution is

______ M

(nearest integer) [Given molar mass of

H₂SO₄ = 98g mol⁻¹

]

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Molarity (M) = % by mass \times 10 \times dMw

Core Logic

Let's assume we have 100 g of the solution. Mass of $H₂SO₄$ $H_2SO_4$ in 100 g solution = $31.4 g$ $31.4\text{ g}$. Moles of $H₂SO₄$ $H_2SO_4$ ( $nsolute$ $n_{\text{solute}}$) = $(31.4)/(98) mol$ $\frac{31.4}{98}\text{ mol}$.

Volume of the solution ( $V$ $V$) can be found using density:

V = Mass of solutionDensity = (100)/(1.25) mL

Step 1: Calculating Molarity

Molarity is defined as moles of solute per liter of solution:

M = nsoluteV(in mL) \times 1000

M = (31.4 / 98)/(100 / 1.25) \times 1000

M = (31.4 \times 1.25)/(98 \times 100) \times 1000

M = (39.25)/(98) \times 10

M = 0.4005 \times 10

M = 4.005 M

Rounding off to the nearest integer gives 4.

Pattern Recognition

Whenever percentage by mass ( $w/w$ $w/w$) and density ( $d$ $d$ in g/mL) are given, use the direct formula: $M = (%(w/w) \times d \times 10)/(Mw)$ . This saves enormous time.

Chapter Mix

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

Q86 jee_main_2024_29_jan_morning Osmotic Pressure

The osmotic pressure of a dilute solution is

7 \times 10⁵ ~Pa

273 ~K

. Osmotic pressure of the same solution at

283 ~K

______ \times 10⁴ Nm⁻²

Numerical Answer. Answer: 72.5 to 73

Solution

Related Formula

$π = C R T$ $\pi = C R T$

where $π$ $\pi$ is osmotic pressure, $C$ $C$ is molar concentration, $R$ $R$ is gas constant, and $T$ $T$ is absolute temperature.

Core Logic

For a given dilute solution, the concentration ( $C$ $C$) and the gas constant ( $R$ $R$) are constant. Therefore, osmotic pressure is directly proportional to the absolute temperature. $π \propto T$ $\pi \propto T$

(π₁)/(T₁) = (π₂)/(T₂)

Step 1: Calculation

Given values: $π₁ = 7 \times 10⁵ Pa = 70 \times 10⁴ Nm⁻²$ $T₁ = 273 K$ $T_1 = 273 \text{ K}$ $T₂ = 283 K$ $T_2 = 283 \text{ K}$

Rearranging for $π₂$ $\pi_2$:

π₂ = (π₁ \cdot T₂)/(T₁)

π₂ = (7 \times 10⁵ \times 283)/(273)

π₂ = (1981 \times 10⁵)/(273)

π₂ = 7.2564 \times 10⁵ Pa

Converting to the requested format ( $\times 10⁴ Nm⁻²$ $\times 10^4 \text{ Nm}^{-2}$):

π₂ = 72.564 \times 10⁴ Nm⁻²

Rounding off yields 72.56 (or 73 depending on required decimal places).

Chapter Mix

Class 12 Chemistry: Solutions

Q72 jee_main_2024_30_january_evening Concentration Terms

If a substance 'A' dissolves in solution of a mixture of 'B' and 'C' with their respective number of moles as

nA

nB

and

nC

, mole fraction of C in the solution is:

A. $(nC)/(nA \times nB \times nC)$ $\frac{n_C}{n_A \times n_B \times n_C}$
B. $(nC)/(nA + nB + nC)$ $\frac{n_C}{n_A + n_B + n_C}$
C. $(nC)/(nA - nB - nC)$ $\frac{n_C}{n_A - n_B - n_C}$
D. $(nB)/(nA + nB)$ $\frac{n_B}{n_A + n_B}$

Solution

Related Formula

χᵢ = nᵢntotal

Core Logic

The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles of all components present in the solution.

Total number of moles in the solution $= nA + nB + nC$ $= n_A + n_B + n_C$

Mole fraction of C ( $χC$ $\chi_C$) = $(nC)/(nA + nB + nC)$ $\frac{n_C}{n_A + n_B + n_C}$

Chapter Mix

Class 12 Chemistry: Solutions

Assume a living cell with 0.9%$0.9\%$ (ω/ω)$(\omega/\omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will:

Solution & Explanation

Related Formula

Core Logic

Step 1: Concentration Determination and Osmosis Profile

Pattern Recognition

Chapter Mix

Reference Study Guides

More Solutions Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Final Resolution

Pattern Recognition

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Solution

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Calculating Molarity

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Calculation

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Solution

Related Formula

Core Logic

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Assume a living cell with $0.9%$ $0.9\%$ $(ω/ω)$ $(\omega/\omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will: