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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Q87 jee_main_2024_31_jan_evening Concentration Terms
The molarity of 1text L orthophosphoric acid (H_3PO_4) having 70\% purity by weight (specific gravity 1.54text g cm^-3) is ________ textM. (Molar mass of H_3PO_4 = 98text g mol^-1)
Numerical Answer. Answer: 11 to 11

Solution

### Related Formula M = frac\% text purity times textdensity times 10textMolar Mass ### Core Logic Specific gravity is numerically equivalent to density in textg/cm^3, so density = 1.54text g/mL. Volume of solution = 1text L = 1000text mL. Mass of solution = textVolume times textDensity = 1000 times 1.54 = 1540text g. ### Step 1: Finding Solute Mass and Molarity Since the purity is 70\% by weight, the mass of H_3PO_4 in the solution is: textMass of H_3PO_4 = 1540 times 0.70 = 1078text g. Moles of H_3PO_4 = frac107898 = 11text moles. Since this is dissolved in 1text L of solution, the Molarity is: M = frac11text moles1text L = 11text M ### Pattern Recognition Shortcut formula directly substitutes the values: M = frac70 times 1.54 times 1098 = frac107898 = 11. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry
Q63 jee_main_2024_31_jan_morning Non-Ideal Solutions
Identify the mixture that shows positive deviations from Raoult's Law
  • A. (CH_3)_2CO + C_6H_5NH_2
  • B. CHCl_3 + C_6H_6
  • C. CHCl_3 + (CH_3)_2CO
  • D. (CH_3)_2CO + CS_2

Solution

### Core Logic (CH_3)_2CO + CS_2 exhibits positive deviations from Raoult's Law because the interactions between acetone and carbon disulphide molecules are weaker than the respective pure component interactions. ### Pattern Recognition Mixtures like Acetone + Aniline or Chloroform + Benzene/Acetone form stronger hydrogen bonds after mixing, showing negative deviation. Acetone + CS_2 or Ethanol + Acetone break existing strong interactions, leading to positive deviation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

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