Q4. Find the osmotic pressure, in pascals, of a solution of 1.0 g of a polymer of molar mass 185 000 g mol−1 in 450 mL of water at 37 °C (NCERT Intext Question 1.12).
n = 1.0/185 000 = 5.41 × 10−6 mol; the solution is very dilute, so its volume is taken as the 450 mL of water (the polymer adds well under 1%), V ≈ 4.50 × 10−4 m3, and T = 310 K; Π = nRT/V = 5.41 × 10−6 × 8.314 × 310/(4.50 × 10−4) ≈ 31 Pa (worked here).
Q5. Assertion: 0.1 mol of glucose and 0.1 mol of urea, each in 1 kg of water, lower its freezing point by the same amount. Reason: depression of freezing point is a colligative property. Are both true, and does the reason explain the assertion?
Both are true within the dilute ideal model, and the reason explains the assertion: both solutions contain the same number of non-dissociating solute particles (NCERT pages 15 and 18).
Q6. Find the freezing point of a 0.20 m aqueous solution of a non-electrolyte (Kf = 1.86 K kg mol−1, water freezing at 273.15 K).
ΔTf = 1.86 × 0.20 = 0.372 K, so Tf = 273.15 − 0.372 = 272.78 K (NCERT Eq. (1.34), page 18).
Q7. Why is osmotic pressure the preferred colligative property for finding the molar masses of proteins?
It is measured near room temperature, where proteins are stable, and it is large even for very dilute solutions (NCERT pages 21–22).
Q8. Using NCERT Eq. (1.38), estimate Kb for water from ΔvapH = 40 790 J mol−1 (NCERT Class 11 Chemistry, Unit 5, page 148), R = 8.314 J mol−1 K−1, M1 = 18 g mol−1 and Tb = 373.15 K.
Kb = 8.314 × 18 × 373.152/(1000 × 40 790) ≈ 0.51 K kg mol−1, close to the 0.52 of NCERT Table 1.3 (worked here).
Q9. A red blood cell placed in a dilute aqueous NaCl solution swells. Is the solution hypertonic or hypotonic in this model?
Hypotonic: water flows into the cell, so its effective concentration of retained particles is lower; in this simple NaCl comparison it is below approximately 0.9% (mass/volume) NaCl (NCERT page 22).
Q10. A 5% (mass/volume) cane sugar solution (342 g mol−1) is isotonic with a 1% (mass/volume) solution of a non-electrolyte X. Find the molar mass of X.
Isotonic non-electrolyte solutions have equal molarities, since Π = CRT (NCERT Eq. (1.39), page 21): 50/342 = 10/M, so M ≈ 68.4 g mol−1 (worked here).