The percentage dissociation of a salt (mathrmMX_3) solution at given temperature (van't Hoff factor texti = 2) is ______ %. (Nearest integer)

Numerical Answer Type:
Enter a numerical value Answer: 33 to 33 +4 marks

Solution & Explanation

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