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Molality of 0.8mathrmM mathrmH_2mathrmSO_4 solution (density 1.06mathrmgcm^-3) is ________ times 10^-3mathrmm.

Numerical Answer Type:
Enter a numerical value Answer: 815 to 815 +4 marks

Solution & Explanation

### 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

Reference Study Guides

Solutions

More Solutions Previous-Year Questions — Page 7

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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