If mathrmA_2mathrmB$\mathrm{A}_2\mathrm{B}$ is 30\%$30\%$ ionised in an aqueous solution, then the value of van't Hoff factor (i) is ________ times 10^-1$\times 10^{-1}$.
Numerical Answer Type:
Enter a numerical valueAnswer: 16 to 16+4 marks
Solution & Explanation
### Related Formula
i = 1 + (y - 1)alpha$i = 1 + (y - 1)\alpha$
### Core Logic
For electrolyte mathrmA_2mathrmB$\mathrm{A}_2\mathrm{B}$ undergoing dissociation :
mathrmA_2B
ightarrow 2mathrmA^+ + mathrmB^2-$\mathrm{A_2B}
ightarrow 2\mathrm{A}^+ + \mathrm{B}^{2-}$
Total count of ions produced per molecule y = 3$y = 3$ .
Given degree of dissociation alpha = 30\% = 0.3$\alpha = 30\% = 0.3$ .
Substituting into the formula :
i = 1 + (3 - 1) cdot 0.3$i = 1 + (3 - 1) \cdot 0.3$i = 1 + 2 cdot 0.3 = 1 + 0.6 = 1.6$i = 1 + 2 \cdot 0.3 = 1 + 0.6 = 1.6$
Expressing the value in the requested format:
1.6 = 16 times 10^-1$1.6 = 16 \times 10^{-1}$
Thus, the integer value to enter is 16.
### Pattern Recognition
Always calculate total stoichiometric species y$y$ carefully before executing the linear factor combination to avoid basic arithmetic errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Solutions
Given below are two statements :
Statement (I): Molal depression constant K_f$K_f$ is given by fracM_lRT_fDelta S_fus$\frac{M_lRT_f}{\Delta S_{fus}}$, where symbols have their usual meaning.
Statement (II): K_f$K_f$ for benzene is less than the K_f$K_f$ for water.
In the light of the above statements, choose the most appropriate answer from the options given below:
A. Statement I is incorrect but Statement II is correct
B. Both Statement I and Statement II are incorrect.
C. Both Statement I and Statement II are correct
D. Statement I is correct but Statement II is incorrect
Solution
### Related Formula
K_f = fracM_1 R T_f^2Delta H_fus = fracM_1 R T_fleft(fracDelta H_fusT_fright) = fracM_1 R T_fDelta S_fus$K_f = \frac{M_1 R T_f^2}{\Delta H_{fus}} = \frac{M_1 R T_f}{\left(\frac{\Delta H_{fus}}{T_f}\right)} = \frac{M_1 R T_f}{\Delta S_{fus}}$
### Core Logic
- **Statement I is correct:** Substituting Delta S_fus = fracDelta H_fusT_f$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$ directly matches the given structural relationship formula.
- **Statement II is incorrect:** Standard cryoscopic constants are:
- For Benzene: K_f approx 5.12 mathrm~^circ C cdot kg cdot mol^-1$K_f \approx 5.12 \mathrm{~^\circ C \cdot kg \cdot mol^{-1}}$
- For Water: K_f approx 1.86 mathrm~^circ C cdot kg cdot mol^-1$K_f \approx 1.86 \mathrm{~^\circ C \cdot kg \cdot mol^{-1}}$
Therefore, K_f$K_f$ for benzene is greater than that of water, making Statement II false.
### Pattern Recognition
Keep numerical benchmarks for common solvent colligative constants (K_b, K_f$K_b, K_f$ for water and benzene) memorized. Benzene has a far lower enthalpy of fusion and a higher freezing point, resulting in a significantly elevated K_f$K_f$ value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Solutions
Q462025Concentration Terms
Sea water, which can be considered as a 6 molar (6 M) solution of NaCl, has a density of 2mathrm~g~mL^-1$2\mathrm{~g~mL}^{-1}$ . The concentration of dissolved oxygen left(mathrmO_2right)$\left(\mathrm{O}_2\right)$ in sea water is 5.8mathrm~ppm$5.8\mathrm{~ppm}$ . Then the concentration of dissolved oxygen left(mathrmO_2right)$\left(\mathrm{O}_2\right)$ in sea water, is mathrmx times 10^-4mathrmm$\mathrm{x} \times 10^{-4}\mathrm{m}$ . mathrmx =$\mathrm{x} =$ _______. (Nearest integer)
Given: Molar mass of NaCl is 58.5mathrm~g~mol^-1$58.5\mathrm{~g~mol}^{-1}$ Molar mass of mathrmO_2$\mathrm{O}_2$ is 32mathrm~g~mol^-1$32\mathrm{~g~mol}^{-1}$
Numerical Answer.Answer: 1.9 to 2.1
Solution
### Related Formula
textppm = fractextmass of solutetextmass of solution times 10^6$\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$textMolality (m) = fractextmoles of solutetextmass of solvent in kg$\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
### Core Logic
1. Consider 1000 mathrm~mL$1000 \mathrm{~mL}$ of seawater solution:
textMass of solution = textVolume times textdensity = 1000 times 2 = 2000 mathrm~g$\text{Mass of solution} = \text{Volume} \times \text{density} = 1000 \times 2 = 2000 \mathrm{~g}$textMass of NaCl = 6 text moles times 58.5 = 351 mathrm~g$\text{Mass of NaCl} = 6 \text{ moles} \times 58.5 = 351 \mathrm{~g}$textMass of solvent (water) = 2000 - 351 = 1649 mathrm~g = 1.649 mathrm~kg$\text{Mass of solvent (water)} = 2000 - 351 = 1649 \mathrm{~g} = 1.649 \mathrm{~kg}$
2. Compute the mass and moles of dissolved O_2$O_2$ using the ppm value:
textppm = 5.8 = fractextmass of O_22000 times 10^6 implies textmass of O_2 = 1.16 times 10^-2 mathrm~g$\text{ppm} = 5.8 = \frac{\text{mass of } O_2}{2000} \times 10^6 \implies \text{mass of } O_2 = 1.16 \times 10^{-2} \mathrm{~g}$textmoles of O_2 = frac1.16 times 10^-232 = 3.625 times 10^-4 text moles$\text{moles of } O_2 = \frac{1.16 \times 10^{-2}}{32} = 3.625 \times 10^{-4} \text{ moles}$
3. Determine the molality (m$m$) of oxygen:
textmolality = frac3.625 times 10^-41.649 approx 2.19 times 10^-4 mathrm~m$\text{molality} = \frac{3.625 \times 10^{-4}}{1.649} \approx 2.19 \times 10^{-4} \mathrm{~m}$
Matching the pattern mathbfx times 10^-4mathrmm$\mathbf{x} \times 10^{-4}\mathrm{m}$, we get mathbfx approx 2.19$\mathbf{x} \approx 2.19$. The nearest integer is **2**.
### Pattern Recognition
For high concentration saline solutions, the mass of the solvent drops significantly below the total mass of the solution. Be careful to subtract the solute weight (351 mathrm~g$351 \mathrm{~g}$) before computing molality.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Solutions
Q262025Reverse Osmosis
XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration c_1$c_1$ and c_2$c_2$ (c_1 > c_2$c_1 > c_2$) mathrmmol~L^-1$\mathrm{mol~L^{-1}}$. For the reverse osmosis to take place identify the correct condition (Here p_1$p_1$ and p_2$p_2$ are pressures applied on chamber 1 and 2):
The diagram illustrates two chambers separated by a membrane XY containing sugar solutions of concentrations c1 and c2.
A.text(B) and (D) only$\text{(B) and (D) only}$
B.text(A) and (D) only$\text{(A) and (D) only}$
C.text(A) and (C) only$\text{(A) and (C) only}$
D.text(C) only$\text{(C) only}$
Solution
### Related Formula
pi = c R T$\pi = c R T$
where pi$\pi$ is the osmotic pressure of the solution.
### Core Logic
Given that c_1 > c_2$c_1 > c_2$, chamber 1 has a higher concentration of solute than chamber 2. Under normal conditions, solvent molecules spontaneously flow from lower concentration (chamber 2) to higher concentration (chamber 1) via osmosis.
To achieve **reverse osmosis**, the solvent must flow in the opposite direction—from chamber 1 to chamber 2. This requires applying an external pressure on the higher concentration side (chamber 1) that exceeds its osmotic pressure pi$\pi$.
textCondition for Reverse Osmosis: p_1 > pi$\text{Condition for Reverse Osmosis: } p_1 > \pi$
Cellophane and parchment paper both act as suitable semi-permeable membranes for this setup. Thus, statements (A) and (C) are correct.
### Pattern Recognition
Reverse osmosis always requires external pressure applied on the concentrated solution side (c_texthigh$c_{\text{high}}$) such that P_textapplied > pi$P_{\text{applied}} > \pi$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Solutions
Q342025Raoult's Law and Liquid-Vapour Composition
Liquid textA$\text{A}$ and textB$\text{B}$ form an ideal solution. The vapour pressure of pure liquids textA$\text{A}$ and textB$\text{B}$ are 350 and 750text mm Hg$750\text{ mm Hg}$ respectively at the same temperature. If textx_textA$\text{x}_{\text{A}}$ and textx_textB$\text{x}_{\text{B}}$ are the mole fraction of textA$\text{A}$ and textB$\text{B}$ in solution while texty_textA$\text{y}_{\text{A}}$ and texty_textB$\text{y}_{\text{B}}$ are the mole fraction of textA$\text{A}$ and textB$\text{B}$ in vapour phase then:
### Related Formula
textNegative Deviation from Raoult's Law implies textMaximum Boiling Azeotrope $\text{Negative Deviation from Raoult's Law} \implies \text{Maximum Boiling Azeotrope} $textPositive Deviation from Raoult's Law implies textMinimum Boiling Azeotrope $\text{Positive Deviation from Raoult's Law} \implies \text{Minimum Boiling Azeotrope} $textIdeal Solution implies Delta Vtextmix = 0 $\text{Ideal Solution} \implies \Delta V{\text{mix}} = 0 $
### Core Logic
Evaluating molecular interaction behaviors:
- (A) Chloroform and Acetone: Form intermolecular hydrogen bonds, showing negative deviation from Raoult's law, which forms a maximum boiling azeotrope
ightarrow$
ightarrow$ (III)
- (B) Ethanol and Water: Exhibit hydrogen bond disruptions, leading to positive deviation, forming a minimum boiling azeotrope
ightarrow$
ightarrow$ (I)
- (C) Benzene and Toluene: Highly structurally similar, forming a near-perfect ideal solution where Delta Vtextmix = 0$\Delta V{\text{mix}} = 0$
ightarrow$
ightarrow$ (IV)
- (D) Acetic acid in benzene: Acetic acid undergoes intermolecular hydrogen bonding inside the non-polar solvent, causing it to dimerize
ightarrow$
ightarrow$ (II)
### Step 1: Alignment Selection
Combining the pairs gives: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
### Pattern Recognition
Liquid properties shortcut: Acetic acid in benzene is a classic textbook example of dimerization (i=0.5$i=0.5$). Benzene-toluene is the most famous ideal solution pair. Recognizing either instantly shortcuts the options list to the correct key.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Solutions
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