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

Q jee_main_2025_29_jan_morning van't Hoff Factor

If

mathrmA_2mathrmB

is

30\%

ionised in an aqueous solution, then the value of van't Hoff factor (i) is ________

times 10^-1

.

Numerical Answer. Answer: 16 to 16

Solution

### Related Formula

i = 1 + (y - 1)alpha

### Core Logic For electrolyte

mathrmA_2mathrmB

undergoing dissociation:

mathrmA_2B rightarrow 2mathrmA^+ + mathrmB^2-

Total count of ions produced per molecule

y = 3

. Given degree of dissociation

alpha = 30\% = 0.3

. Substituting into the formula:

i = 1 + (3 - 1) cdot 0.3

i = 1 + 2 cdot 0.3 = 1 + 0.6 = 1.6

Expressing the value in the requested format:

1.6 = 16 times 10^-1

Thus, the integer value to enter is 16. ### Pattern Recognition Always calculate total stoichiometric species

y

carefully before executing the linear factor combination to avoid basic arithmetic errors. ### Chapter Mix Class 12 Chemistry: Solutions

Q75 jee_main_2024_01_february_morning Abnormal Molar Masses

We have three aqueous solutions of

NaCl

labelled as 'A', 'B' and 'C' with concentration

0.1 mathrm~M

,

0.01 mathrm~M

&

0.001 mathrm~M

, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order.

A. $i_A < i_B < i_C$ $i_A < i_B < i_C$
B. $i_A < i_C < i_B$ $i_A < i_C < i_B$
C. $i_A = i_B = i_C$ $i_A = i_B = i_C$
D. $i_A > i_B > i_C$ $i_A > i_B > i_C$

Solution

### Core Logic For a strong electrolyte like

NaCl

, the theoretical van 't Hoff factor

i_theory

is 2 (

Na^+

and

Cl^-

). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed

i

. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed

i

approaches the theoretical value. ### Step 1: Correlate Concentration with

i

Higher concentration

implies

more ion-pairing

implies

lower

i

. Given concentrations:

A = 0.1 mathrm~M

(highest concentration)

B = 0.01 mathrm~M

C = 0.001 mathrm~M

(most dilute) Therefore, the actual

i

values follow the reverse order of concentration:

i_A (0.1 mathrm~M) < i_B (0.01 mathrm~M) < i_C (0.001 mathrm~M)

. ### Execution

Salt	Values of i (for different conc. of a Salt)
Salt	0.1 M	0.01 M	0.001 M
NaCl	1.87	1.94	1.97

i

approaches 2 as the solution becomes very dilute. ### Pattern Recognition For strong electrolytes, effective dissociation (and thus

i

) increases as dilution increases because ions interfere with each other less. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

Q84 jee_main_2024_29_january_evening Interconversion of Concentration Terms

Molality of

0.8mathrmM

mathrmH_2mathrmSO_4

solution (density

1.06mathrmgcm^-3

) is ________

times 10^-3mathrmm

.

Numerical Answer. Answer: 815 to 815

Solution

### Related Formula

m = fracM times 1000, (1000 times d) - (M times M_B)

where,

M

= Molarity =

0.8text M

d

= Density =

1.06text g/cm^3

M_B

= Molar mass of solute (

H_2SO_4

) =

98text g/mol

### Core Logic Substituting the given values into the equation:

m = frac0.8 times 1000, (1000 times 1.06) - (0.8 times 98)

Calculating the denominator parameters:

textDenominator = 1060 - 78.4 = 981.6text g

### Step 1: Final Resolution Solving for molality:

m = frac800, 981.6 approx 0.815text m = 815 times 10^-3text 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. ### Evaluation Rubric / Model Answer null ### 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

interactions are stronger than

A-A

or

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

rightarrow

Vapour Pressure drops

rightarrow

Boiling Point rises inversely. ### Chapter Mix Class 12 Chemistry: Solutions

Q85 jee_main_2024_29_jan_morning Concentration Terms

A solution of

mathrmH_2mathrmSO_4

is

31.4\%

mathrmH_2mathrmSO_4

by mass and has a density of

1.25mathrmg / mL

. The molarity of the

mathrmH_2mathrmSO_4

solution is

\_\_\_\_\_\_ mathrmM

(nearest integer) [Given molar mass of

mathrmH_2mathrmSO_4 = 98mathrmg mol^-1

]

Numerical Answer. Answer: 4 to 4

Solution

### Related Formula

textMolarity (M) = frac\% text by mass times 10 times dM_w

### Core Logic Let's assume we have 100 g of the solution. Mass of

H_2SO_4

in 100 g solution =

31.4text g

. Moles of

H_2SO_4

(

n_textsolute

) =

frac31.498text mol

. Volume of the solution (

V

) can be found using density:

V = fractextMass of solutiontextDensity = frac1001.25text mL

### Step 1: Calculating Molarity Molarity is defined as moles of solute per liter of solution:

M = fracn_textsoluteV(textin mL) times 1000

M = frac31.4 / 98100 / 1.25 times 1000

M = frac31.4 times 1.2598 times 100 times 1000

M = frac39.2598 times 10

M = 0.4005 times 10

M = 4.005text M

Rounding off to the nearest integer gives 4. ### Pattern Recognition Whenever percentage by mass (

w/w

) and density (

d

in g/mL) are given, use the direct formula:

M = frac\%(w/w) times d times 10M_w

. This saves enormous time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry

Assume a living cell with $0.9\%$ $0.9\%$ $(omega/omega)$ $(\omega/\omega)$ 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

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Assume a living cell with 0.9\%$0.9\%$ (omega/omega)$(\omega/\omega)$ 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

Reference Study Guides

More Solutions Previous-Year Questions — Page 5

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Assume a living cell with $0.9\%$ $0.9\%$ $(omega/omega)$ $(\omega/\omega)$ 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: