PYQ Vault

Van't Hoff factor from a measured colligative property

This type of question, asked 10 times: 2 May 2023 S2, 10 May 2023 S1, 11 May 2023 S1, 16 May 2023 S1, 12 May 2024 S2, 13 May 2024 S2, 23 Apr 2025 S1, 17 Apr 2026 S1, 13 May 2026 S1, 15 May 2026 S2. All 10 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2 May 2023 S2

    91
    Calculate van't Hoff factor of 0.2 m aqueous solution of an electrolyte if it freezes at −0.7 K-0.7\,\text{K}. Kf=1.86 K kg mol−1K_f = 1.86\,\text{K kg mol}^{-1}

    Tap an option to check your answer.

  2. Question 2

    10 May 2023 S1

    86
    0.2 molal aqueous solution of KCl freezes at −0.680∘C-0.680^\circ\text{C}. Calculate van't Hoff factor for this solution. Kf=1.86 K kg mol−1K_f=1.86\,\text{K kg mol}^{-1}

    Tap an option to check your answer.

  3. Question 3

    11 May 2023 S1

    56
    Calculate van't Hoff factor (i) of 0.2 m aqueous solution of an electrolyte if it freezes at −0.660-0.660 K. Kf=1.84K_f = 1.84 K kg mol−1^{-1}

    Tap an option to check your answer.

  4. Question 4

    16 May 2023 S1

    94
    Calculate van't Hoff factor of 0.15 M solution of electrolyte if it freezes at −0.5 K-0.5\,\text{K}. Kf=1.86 K kg mol−1K_f = 1.86\,\text{K kg mol}^{-1}

    Tap an option to check your answer.

  5. Question 5

    12 May 2024 S2

    96
    If 0.15 m aqueous solution of KCl freezes at −0.51°C-0.51°C, calculate van't Hoff factor of KCl (cryoscopic constant of water is 1.861.86 K kg mol⁻¹)

    Tap an option to check your answer.

  6. Question 6

    13 May 2024 S2

    58
    If 0.01 m aqueous solution of an electrolyte freezes at −0.056 K-0.056\ \text{K}. Calculate Van't Hoff factor for an electrolyte (cryoscopic constant of water =1.86 K kg mol−1=1.86\ \text{K kg mol}^{-1})

    Tap an option to check your answer.

  7. Question 7

    23 Apr 2025 S1

    85
    Calculate van't Hoff factor of aqueous solution of 0.18 m electrolyte that freezes at −0.54∘C-0.54^{\circ}C. ( KfK_{f} for solvent =1.86 K kg mol−1= 1.86\text{ }K\text{ }kg{\text{ }mol}^{- 1} )

    Tap an option to check your answer.

  8. Question 8

    17 Apr 2026 S1

    22
    Calculate van't Hoff factor for aqueous solution of 0.02 m formic acid if it freezes at -0.045 ∘C^\circ\mathrm{C}. [ KfK_f = 1.86 K kg mol−1\mathrm{mol^{-1}} and freezing point of water = 0 ∘C0\ ^\circ\mathrm{C} ]

    Tap an option to check your answer.

  9. Question 9

    13 May 2026 S1

    21
    A 0.10.1 molal aqueous solution of K3[Fe(CN)6]\mathrm{K_3[Fe(CN)_6]} freezes at −0.85 ∘-0.85\ ^\circC. If KfK_f for water is 1.86 K kg mol−11.86\ \mathrm{K\,kg\,mol^{-1}}, what is the van't Hoff factor for this solution?

    Tap an option to check your answer.

  10. Question 10

    15 May 2026 S2

    22
    Calculate van't Hoff factor for CaCl2\mathrm{CaCl_2}, if osmotic pressure of CaCl2\mathrm{CaCl_2} and urea solutions of same concentrations at same temperature are 0.650.65 and 0.270.27 atm respectively.

    Tap an option to check your answer.