When 1~mathrmg$1~\mathrm{g}$ each of compounds AB and mathrmAB_2$\mathrm{AB}_2$ are dissolved in 15~mathrmg$15~\mathrm{g}$ of water separately, they increased the boiling point of water by 2.7~mathrmK$2.7~\mathrm{K}$ and 1.5~mathrmK$1.5~\mathrm{K}$ respectively. The atomic mass of A (in amu) is times 10^-1$\times 10^{-1}$ (Nearest integer) (Given : Molal boiling point elevation constant is 0.5~mathrmK~kg~mol^-1$0.5~\mathrm{K~kg~mol^{-1}}$)

### Related Formula Delta T_b = K_b times m$$\Delta T_b = K_b \times m$$ m = fractextWeight of solute (g)textMolar mass of solute times frac1000textWeight of solvent (g)$$m = \frac{\text{Weight of solute (g)}}{\text{Molar mass of solute}} \times \frac{1000}{\text{Weight of solvent (g)}}$$ ### Core Logic For compound PQ: (Delta mathrmT_b)_mathrmPQ = mathrmK_b cdot m$$(\Delta \mathrm{T_b})_{\mathrm{PQ}} = \mathrm{K_b} \cdot m$$ 1.176 = 5 times frac1mathrmM_1 times frac100050$$1.176 = 5 \times \frac{1}{\mathrm{M}_1} \times \frac{1000}{50}$$ mathrmM_1 = frac5 times 201.176 = 85.03 text g/mol$$\mathrm{M}_1 = \frac{5 \times 20}{1.176} = 85.03 \text{ g/mol}$$ For compound PQ_2$PQ_2$: (Delta mathrmT_b)_mathrmPQ_2 = 5 times frac1mathrmM_2 times frac100050 = 0.689$$(\Delta \mathrm{T_b})_{\mathrm{PQ}_2} = 5 \times \frac{1}{\mathrm{M}_2} \times \frac{1000}{50} = 0.689$$ mathrmM_2 = frac5 times 200.689 = 145.13 text g/mol$$\mathrm{M}_2 = \frac{5 \times 20}{0.689} = 145.13 \text{ g/mol}$$ Let molar mass of P & Q be mathrmM_P$\mathrm{M}_P$ and mathrmM_Q$\mathrm{M}_Q$ respectively: mathrmM_P + mathrmM_Q = 85.03 quad text--- (1)$$\mathrm{M}_P + \mathrm{M}_Q = 85.03 \quad \text{--- (1)}$$ mathrmM_P + 2mathrmM_Q = 145.13 quad text--- (2)$$\mathrm{M}_P + 2\mathrm{M}_Q = 145.13 \quad \text{--- (2)}$$ Subtracting (1) from (2): mathrmM_Q = 145.13 - 85.03 = 60.1 approx 60 text g/mol$$\mathrm{M}_Q = 145.13 - 85.03 = 60.1 \approx 60 \text{ g/mol}$$ Substituting back into (1): mathrmM_P + 60.1 = 85.03 implies mathrmM_P = 24.93 approx 25 text g/mol$$\mathrm{M}_P + 60.1 = 85.03 \implies \mathrm{M}_P = 24.93 \approx 25 \text{ g/mol}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

### 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 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