Class 12 Chemistry NCERT Chapter 1: Higher Order Thinking Skills (HOTS) Questions
These HOTS class 12 chemistry chapter 1 solutions question answer are designed to test students understanding and application of concepts. These NCERT Solutions help in developing critical thinking skills and prepare students for complex questions in competitive exams.
Question 1. $\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$
The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20^{\circ} \mathrm{C}$. The dissociation constant for the acid is
Given :
$\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right)=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, molality $\equiv$ molarity
(1). $1.38 \times 10^{-3}$
(2). $1.1 \times 10^{-2}$
(3). $1.90 \times 10^{-3}$
(4). $1.89 \times 10^{-1}$
Answer:
$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\ & \mathrm{i}=\frac{\Delta \mathrm{T}_f}{\mathrm{~K}_{\mathrm{f}} \cdot \mathrm{m}} \\ & \mathrm{i}=\frac{0.20}{1.8 \times 0.1}=1.11 \\ & \mathrm{i}=1.11 \\ & \alpha=\frac{i-1}{\mathrm{n}-1}(\text { for } \mathrm{HA}, \mathrm{n}=2) \\ & \alpha=\frac{1.11-1}{1}=0.11 \\ & \mathrm{~K}_{\mathrm{a}}=\frac{c \alpha^2}{1-\alpha}=\frac{0.1 \times(0.11)^2}{1-0.11}=1.38 \times 10^{-3}\end{aligned}$
Hence, the correct answer is option (1).
Question 2. Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes ?
(1) $\mathrm{H}_2 \mathrm{O}+\mathrm{CH}_3 \mathrm{COC}_2 \mathrm{H}_5$
(2) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
(3) $\mathrm{CS}_2+\mathrm{CH}_3 \mathrm{COCH}_3$
(4) $\mathrm{CH}_3 \mathrm{OH}+\mathrm{CHCl}_3$
Answer:
The solution showing positive deviation from Raoult's law will form a minimum boiling azeotrope. Phenol + Aniline shows negative deviation, so they will not form a minimum boiling azeotrope.
A binary mixture of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ exhibits negative deviation from Raoult's law. This means that the vapor pressure of the solution is lower than the vapor pressures of the pure components, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$. Consequently, the boiling point of the solution is higher than the boiling points of the pure substances. Therefore, this mixture forms a maximum boiling azeotrope.
Hence, the correct answer is option (2).
Question 3. Given below are two statements :
Statement (I) : Molal depression constant $\mathrm{K}_{\mathrm{f}}$ is given by $\frac{M_1 R T_f}{\Delta S_{f u s}}$, where symbols have their usual meaning.
Statement (II) : $\mathrm{K}_{\mathrm{f}}$ for benzene is less than the $\mathrm{K}_f$ for water.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are incorrect
(3) Both Statement I and Statement II are correct
(4) Statement I is correct but Statement II is incorrect
Answer: 
Statement-I
Molar depression constant $\mathrm{k}_f=\frac{\mathrm{M}_{\mathrm{i}} \mathrm{RT}_{\mathrm{f}}^2}{\Delta \mathrm{H}_{\text {fius }}}$
$\begin{aligned}
& \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_{\mathrm{f}}}{\left[\frac{\Delta \mathrm{H}_{\text {fis }}}{\mathrm{T}_{\mathrm{f}}}\right]} \\
& \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\text {fus }}}
\end{aligned}$
Hence statement-I is correct
but $\mathrm{k}_{\mathrm{f}}$ for benzene $=5.12 \frac{{ }^{\circ} \mathrm{C}}{\text { molal }}$
$\mathrm{k}_{\mathrm{f}}$ for water $=1.86 \frac{{ }^{\circ} \mathrm{C}}{\text { molal }}$ Hence statement- II is incorrect
Hence, the correct answer is option (4).
Question 4. 20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is ________ M. (Nearest Integer value)
(Given : $\mathrm{Na}=23, \mathrm{I}=127, \mathrm{Ag}=108, \mathrm{~N}=14$, $\left.\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}\right)$
Answer: 
Let molarity of Nal solution be $\times \mathrm{M}$
$\mathrm{NaI}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgI}+\mathrm{NaNO}_3$
This is a 1:1 molar reaction — 1 mole of NaI gives 1 mole of AgI.
Moles of AgI $=\frac{4.74}{235} \approx 0.02017 \mathrm{~mol}$
So, moles of Nal = moles of Agl = $\mathbf{0 . 0 2 0 1 7 ~ m o l}$
Step 3: Use molarity formula
$$
\text { Molarity }=\frac{\text { moles }}{\text { volume in } \mathrm{L}}=\frac{0.02017}{0.020} \approx 1.0085 \mathrm{M}
$$
Hence, the answer is 1.
Question 5. Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $\mathrm{x}_{\mathrm{A}}$ and $x_B$ are the mole fraction of $A$ and $B$ in solution while $y_A$ and $y_B$ are the mole fraction of $A$ and $B$ in vapour phase then :
(1) $\frac{x_A}{x_B}<\frac{y_A}{y_B}$
(2) $\frac{x_A}{x_B}=\frac{y_A}{y_B}$
(3) $\frac{x_A}{x_B}>\frac{y_A}{y_B}$
(4) $\left(\mathrm{x}_{\mathrm{A}}-\mathrm{y}_{\mathrm{A}}\right)<\left(\mathrm{x}_{\mathrm{B}}-\mathrm{y}_{\mathrm{B}}\right)$
Answer: 
$\begin{array}{ll}P_A^0=350 \mathrm{~mm} \mathrm{Hg} & \mathrm{P}_B^0=750 \mathrm{~mm} \mathrm{Hg} \\ \mathrm{y}_A=\frac{350 \mathrm{x}_A}{\mathrm{P}_{\mathrm{T}}} & \mathrm{y}_B=\frac{350 \mathrm{x}_B}{\mathrm{P}_{\mathrm{T}}}\end{array}$
$\begin{aligned} & \frac{y_A}{y_B}=\frac{350 x_A}{750 x_B} \\ & \Rightarrow \frac{y_A}{y_B}<\frac{x_A}{x_B}\end{aligned}$
Hence, the correct answer is option (3).
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Approaches to Solve Problems of Chapter 1 Chemistry Class 12
Sometimes, problems related to class 12 chemistry chapter 1 solutions seem difficult, but once we understand the basic formulas, it becomes very easy to solve all the questions related to Solutions. We can follow the steps given below to solve the questions based on the Solutions:
1. The first step is to identify the solute and solvent and learn to classify the solution based on the states of components like solid in liquid, gas in liquid, etc. Also, learn to apply the concentration formulas
Molarity (M), Molality (m), Mole fraction (χ), Mass %, Volume %
(i). Molarity
$M=\frac{\text { moles of solute }}{\text { volume of solution in liters }}$
(ii). Molality
$m=\frac{\text { moles of solute }}{\text { mass of solvent in } \mathrm{kg}}$
(iii). Mole Fraction
$X_A=\frac{\text { moles of component } A}{\text { total moles of solution }}$
2. Important formulas from chapter 1 solutions class 12 chemistry are given below. These formulas help students to solve questions easily.
Henry's Law: $P=k_H \cdot x$ for gas solubility.
Raoult's Law- For ideal solutions, use $p_A=x_A \cdot p_A^0$.
Apply to vapor pressure problems.
3. Identify which property is involved ( $\Delta \mathrm{Tf}, \Delta \mathrm{Tb}, \pi, \Delta \mathrm{P}$ ).
Use-
(i). Elevation in Boiling Point : $\Delta T_b=i \cdot K_b \cdot m$
(ii). Depression in Freezing Point: $\Delta T_f=i \cdot K_f \cdot m$
(iii). Osmotic Pressure: $\pi=i \cdot C \cdot R \cdot T$
4. Learn to determine if solute dissociates or associates. Adjust formulas using ? for non-ideal cases.
5. While solving questions from class 12 chemistry chapter 1 solutions the numericals, keep units consistent and understand what each symbol represents. You can also draw diagrams for clarity also focus on ideal vs non-ideal solutions, positive/negative deviations, and azeotropes. Students can also access the NCERT exemplar solutions for better understanding.