Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ_2. When 1 g of PQ is dissolved in 50 g of solvent ‘A’. Delta T_b was 1.176 K while when 1 g of PQ_2 is dissolved in 50 g of solvent ‘A’, Delta T_b was 0.689 K. (K_b of ‘A’ = 5text K kg mol^-1). The molar masses of elements P and Q (in textg mol^-1) respectively, are :

Solution & Explanation

### Related Formula Delta T_b = K_b times m m = fractextWeight of solute (g)textMolar mass of solute times frac1000textWeight of solvent (g) ### Core Logic For compound PQ: (Delta mathrmT_b)_mathrmPQ = mathrmK_b cdot m 1.176 = 5 times frac1mathrmM_1 times frac100050 mathrmM_1 = frac5 times 201.176 = 85.03 text g/mol For compound PQ_2: (Delta mathrmT_b)_mathrmPQ_2 = 5 times frac1mathrmM_2 times frac100050 = 0.689 mathrmM_2 = frac5 times 200.689 = 145.13 text g/mol Let molar mass of P & Q be mathrmM_P and mathrmM_Q respectively: mathrmM_P + mathrmM_Q = 85.03 quad text--- (1) mathrmM_P + 2mathrmM_Q = 145.13 quad text--- (2) Subtracting (1) from (2): mathrmM_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 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

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Q49 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 ightarrow 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. ### Evaluation Rubric / Model Answer null ### 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
  • B. i_A < i_C < i_B
  • C. i_A = i_B = i_C
  • D. 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
SaltValues of i (for different conc. of a Salt)
0.1 M0.01 M0.001 M
NaCl1.871.941.97
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

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