PYQSide ExamsMy practice

Solutions · Question 38

JEE Main 2025 (Online) 28th January Morning ShiftMCQEasy📈 Topic in 6% papers+4−1
1

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol-1) and the decrease in freezing point is 0.40 K?

  1. 4.43 K kg mol-1
  2. 3.72 K kg mol-1
  3. 5.12 K kg mol-1
  4. 1.86 K kg mol-1

Answer

(3) 5.12 K kg mol-1

Solution

Tf = Kf m, where molality m = wB 1000MB wA

m = 1 1000256 50 = 100012800 = 0.078125 mol kg-1

Kf = Tfm = 0.400.078125

= 5.12 K kg mol-1

Answer: (3)

JEE Main 2025 (Online) 28th January Morning Shift

Keep practising — Solutions

2

Assume a living cell with 0.9% (w/w) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will :

  1. shrink since solution is 0.45% (w/w) as a result of association of glucose molecules (due to hydrogen bonding)
  2. shrink since solution is 0.5% (w/w)
  3. show no change in volume since solution is 0.9% (w/w)
  4. swell up since solution is 1% (w/w)
JEE Main 2025 (Online) 28th January Evening Shift

All Solutions PYQs →

PDFDownload Solutions PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All JEE Main Chemistry chapters
Physical ChemistrySome Basic Concepts of ChemistryStructure of AtomStates of MatterThermodynamicsEquilibriumRedox ReactionsSolutionsElectrochemistryChemical Kinetics
Inorganic ChemistryClassification of Elements and Periodicity in PropertiesChemical Bonding and Molecular Structurep-Block Elements (Group 13 & 14)p-Block Elements (Group 15–18)d- and f-Block ElementsCoordination Compounds
Organic ChemistryOrganic Chemistry – Some Basic Principles and TechniquesIsomerismHydrocarbonsHaloalkanes and HaloarenesAlcohols, Phenols and EthersAldehydes, Ketones and Carboxylic AcidsAminesBiomoleculesPrinciples Related to Practical Chemistry