### Related Formula textMolarity (M) = fracn_textsoluteV_textsolution (mathrmL)$\text{Molarity } (M) = \frac{n_{\text{solute}}}{V_{\text{solution}} (\mathrm{L})}$ ### Core Logic Since solubility of mathrmNaCl$\mathrm{NaCl}$ remains unchanged, the number of dissolved moles of mathrmNaCl$\mathrm{NaCl}$ solute (n_textsolute$n_{\text{solute}}$) remains strictly constant. Thus, molarity M$M$ is strictly dependent on the volume of water (solvent) as temperature changes: M propto frac1V_textsolution$M \propto \frac{1}{V_{\text{solution}}}$ ### Step 1: Understand Water's Anomalous Expansion Water exhibits unique anomalous density behavior near freezing: - From 1^circmathrmC$1^{\circ}\mathrm{C}$ to 4^circmathrmC$4^{\circ}\mathrm{C}$, the density of water **increases** to a maximum. This contraction means the volume (V$V$) of water **decreases**. - From 4^circmathrmC$4^{\circ}\mathrm{C}$ to 25^circmathrmC$25^{\circ}\mathrm{C}$, the density of water **decreases** due to standard thermal expansion. Consequently, the volume (V$V$) **increases**. ### Step 2: Relate Volume to Molarity Because volume is in the denominator of the molarity equation: - From 1^circmathrmC$1^{\circ}\mathrm{C}$ to 4^circmathrmC$4^{\circ}\mathrm{C}$: Volume decreases implies$\implies$ Molarity **increases**. - At 4^circmathrmC$4^{\circ}\mathrm{C}$: Volume is minimum implies$\implies$ Molarity reaches a **maximum**. - From 4^circmathrmC$4^{\circ}\mathrm{C}$ to 25^circmathrmC$25^{\circ}\mathrm{C}$: Volume increases implies$\implies$ Molarity **decreases**. This behavior is perfectly represented by **Plot (2)**, which features a distinct peak around 4^circmathrmC$4^{\circ}\mathrm{C}$. ### Pattern Recognition Water is at its densest (and occupies minimum volume) at exactly 3.98^circmathrmC$3.98^\circ\mathrm{C}$ (4^circmathrmC$4^\circ\mathrm{C}$). Any concentration unit based on volume (such as Molarity or Normality) will reach a corresponding maximum at this temperature. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Delta T_mathrmb = K_mathrmb cdot m = K_mathrmb cdot left( fracw_textsoluteM_textsolute cdot frac1000w_textsolvent right)$\Delta T_{\mathrm{b}} = K_{\mathrm{b}} \cdot m = K_{\mathrm{b}} \cdot \left( \frac{w_{\text{solute}}}{M_{\text{solute}}} \cdot \frac{1000}{w_{\text{solvent}}} \right)$ ### Core Logic Both dissolved compounds are non-electrolytes (van 't Hoff factor i = 1$i = 1$). We calculate the molar masses of mathrmAB$\mathrm{AB}$ and mathrmAB_2$\mathrm{AB_2}$ individually, then solve for the individual atomic masses of elements A and B. ### Step 1: Determine Molar Mass of AB Given Delta T_mathrmb = 2.7~mathrmK$\Delta T_{\mathrm{b}} = 2.7~\mathrm{K}$, w_textsolute = 1~mathrmg$w_{\text{solute}} = 1~\mathrm{g}$, w_textsolvent = 15~mathrmg$w_{\text{solvent}} = 15~\mathrm{g}$, and K_mathrmb = 0.5~mathrmK~kg~mol^-1$K_{\mathrm{b}} = 0.5~\mathrm{K~kg~mol^{-1}}$: 2.7 = 0.5 times frac1M_mathrmAB times frac100015$2.7 = 0.5 \times \frac{1}{M_{\mathrm{AB}}} \times \frac{1000}{15}$ M_mathrmAB = frac0.5 times 100015 times 2.7 = frac50040.5 approx 12.3457~mathrmg~mol^-1$M_{\mathrm{AB}} = \frac{0.5 \times 1000}{15 \times 2.7} = \frac{500}{40.5} \approx 12.3457~\mathrm{g~mol^{-1}}$ ### Step 2: Determine Molar Mass of AB_2 Given Delta T_mathrmb = 1.5~mathrmK$\Delta T_{\mathrm{b}} = 1.5~\mathrm{K}$, w_textsolute = 1~mathrmg$w_{\text{solute}} = 1~\mathrm{g}$, w_textsolvent = 15~mathrmg$w_{\text{solvent}} = 15~\mathrm{g}$, and K_mathrmb = 0.5$K_{\mathrm{b}} = 0.5$: 1.5 = 0.5 times frac1M_mathrmAB_2 times frac100015$1.5 = 0.5 \times \frac{1}{M_{\mathrm{AB_2}}} \times \frac{1000}{15}$ M_mathrmAB_2 = frac0.5 times 100015 times 1.5 = frac50022.5 approx 22.2222~mathrmg~mol^-1$M_{\mathrm{AB_2}} = \frac{0.5 \times 1000}{15 \times 1.5} = \frac{500}{22.5} \approx 22.2222~\mathrm{g~mol^{-1}}$ ### Step 3: Solve for Atomic Mass of A Let the atomic masses of elements A and B be a$a$ and b$b$ respectively: a + b = 12.3457 quad text--- (1)$a + b = 12.3457 \quad \text{--- (1)}$ a + 2b = 22.2222 quad text--- (2)$a + 2b = 22.2222 \quad \text{--- (2)}$ Subtracting equation (1) from (2): b = 22.2222 - 12.3457 = 9.8765~mathrmamu$b = 22.2222 - 12.3457 = 9.8765~\mathrm{amu}$ Substituting b$b$ back into equation (1): a = 12.3457 - 9.8765 = 2.4692~mathrmamu$a = 12.3457 - 9.8765 = 2.4692~\mathrm{amu}$ Expressing a$a$ in the requested format (times 10^-1$\times 10^{-1}$): a = 24.692 times 10^-1 approx 25 times 10^-1$a = 24.692 \times 10^{-1} \approx 25 \times 10^{-1}$ ### Pattern Recognition Mathematical consistency checks: Since mathrmAB_2$\mathrm{AB_2}$ has more atoms than mathrmAB$\mathrm{AB}$ of identical constituent mass, its molar mass should be higher, leading to a smaller elevation of boiling point for the same mass dissolved, which perfectly matches our 2.7~mathrmK rightarrow 1.5~mathrmK$2.7~\mathrm{K} \rightarrow 1.5~\mathrm{K}$ change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Henry's Law formula connects partial pressure to solubility component: p = K_mathrmH cdot x$p = K_{\mathrm{H}} \cdot x$ ### Core Logic As temperature increases, gas dissolution is typically exothermic, meaning solubility initially drops, causing the Henry's constant K_mathrmH$K_{\mathrm{H}}$ to curve upward dynamically before varying at extreme points. For standard non-reactive noble/molecular gases at regular ranges, the magnitude order follows: K_mathrmH(mathrmHe) > K_mathrmH(mathrmN_2) > K_mathrmH(mathrmCH_4)$K_{\mathrm{H}}(\mathrm{He}) > K_{\mathrm{H}}(\mathrm{N_2}) > K_{\mathrm{H}}(\mathrm{CH_4})$ ### Step 1: Selection Graph (4) illustrates the correct relative order and curved profile properly across the given temperature frame. ### Pattern Recognition Higher K_mathrmH$K_{\mathrm{H}}$ value implies lower solubility of that gas at a given pressure. Helium is notoriously insoluble in water compared to organic or polarizable molecules like methane, hence its plot line must live at the top. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula Raoult's Law for a multi-component solution mixtures: P_mathrmS = P_mathrmA^0 cdot X_mathrmA + P_mathrmB^0 cdot X_mathrmB$P_{\mathrm{S}} = P_{\mathrm{A}}^0 \cdot X_{\mathrm{A}} + P_{\mathrm{B}}^0 \cdot X_{\mathrm{B}}$ ### Core Logic Let's determine the mole fractions first based on the input molar amounts: X_mathrmA = frac11+3 = frac14, quad X_mathrmB = frac31+3 = frac34$X_{\mathrm{A}} = \frac{1}{1+3} = \frac{1}{4}, \quad X_{\mathrm{B}} = \frac{3}{1+3} = \frac{3}{4}$ Substitute the known properties into Raoult's equation block: 500 = 200 times frac14 + P_mathrmB^0 times frac34$500 = 200 \times \frac{1}{4} + P_{\mathrm{B}}^0 \times \frac{3}{4}$ 500 = 50 + P_mathrmB^0 times frac34$500 = 50 + P_{\mathrm{B}}^0 \times \frac{3}{4}$ 450 = P_mathrmB^0 times frac34 implies P_mathrmB^0 = 600 mathrm~mmHg$450 = P_{\mathrm{B}}^0 \times \frac{3}{4} \implies P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$ Comparing pure state components pressures: P_mathrmA^0 = 200 mathrm~mmHg$P_{\mathrm{A}}^0 = 200 \mathrm{~mmHg}$ and P_mathrmB^0 = 600 mathrm~mmHg$P_{\mathrm{B}}^0 = 600 \mathrm{~mmHg}$. Lower vapor pressure indicates stronger intermolecular cohesion, making A the least volatile component. ### Step 1: Finalization Thus, the vapour pressure of pure B is 600mathrm~mmHg$600\mathrm{~mmHg}$ and the least volatile chemical is A. ### Pattern Recognition Volatility is directly proportional to pure vapor pressure (P^0$P^0$). Don't mix up 'least volatile' with 'lowest mole fraction'—always evaluate based solely on the isolated values of P^0$P^0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### Related Formula i = 1 + (n - 1)alpha$i = 1 + (n - 1)\alpha$ ### Core Logic For the salt mathrmMX_3$\mathrm{MX}_3$: mathrmMX_3 rightarrow mathrmM^3+ + 3mathrmX^-$\mathrm{MX}_3 \rightarrow \mathrm{M}^{3+} + 3\mathrm{X}^-$ - Number of ions formed per formula unit, n = 1 + 3 = 4$n = 1 + 3 = 4$. Given i = 2$i = 2$, let's find the degree of dissociation (alpha$\alpha$): i = 1 + (4 - 1)alpha$i = 1 + (4 - 1)\alpha$ 2 = 1 + 3alpha implies 3alpha = 1 implies alpha = frac13 approx 0.3333$2 = 1 + 3\alpha \implies 3\alpha = 1 \implies \alpha = \frac{1}{3} \approx 0.3333$ Percentage dissociation: \% text dissociation = alpha times 100 = 33.33 \% approx 33 \%$\% \text{ dissociation} = \alpha \times 100 = 33.33 \% \approx 33 \%$ ### Pattern Recognition Shortcut: For MX_3$MX_3$ dissociating into 4$4$ ions, alpha = (i - 1) / 3$\alpha = (i - 1) / 3$. Since i = 2$i = 2$, alpha = 1/3 = 33\%$\alpha = 1/3 = 33\%$ directly. ### Evaluation Rubric / Model Answer Simple formula application verifying the degree of dissociation to obtain a precise 33$33$ percent. ### Chapter Mix Class 12 Chemistry: Solutions