14. A U-shaped tube with constant cross-section is separated into two halves by a
semi-permeable membrane at its midpoint, which allows water to pass freely but not
glucose. Initially there is 1 litre of water in each arm. An amount of glucose is added
to the left arm, causing the water to rise in that arm. Which of the following do you
need to know to calculate the final height difference of water between the two arms?
a. The amount of glucose added
b. The cross-sectional area of the tube
c. Both (a) and (b)
d. Neither (a) nor (b)
Introduction
In U-shaped tube osmosis experiments, adding glucose to one arm separated by a semi-permeable membrane causes water to move osmotically, creating a height difference between arms. Equilibrium occurs when hydrostatic pressure balances osmotic pressure from glucose. This CSIR NET-style question tests understanding of osmotic pressure formula π = CRT, where height difference h relates via ρgh = π.
Osmosis Mechanism
Water moves from pure water arm (right) to glucose solution arm (left) due to higher osmotic pressure π in the left, which depends on glucose concentration C (moles/volume). Osmotic pressure drives solvent flow until hydrostatic pressure ρgh (density ρ, gravity g, height h) equals π. Final h = π/(ρg) uses constants ρ ≈ 1000 kg/m³ and g ≈ 9.8 m/s², independent of tube geometry for constant cross-section.
Option Analysis
a. The amount of glucose added
Amount determines moles n, and since initial volume is 1 L per arm, C = n/1 L gives π = (n/1)RT. Thus, amount is essential for C and π, hence h.
b. The cross-sectional area of the tube
Constant cross-section means height rise corresponds directly to volume transferred, but equilibrium h depends only on pressure balance ρgh = π, not area, as pressure is independent of area (like uniform hydrostatics). Area cancels out in derivations.
c. Both (a) and (b)
Incorrect, as area is unnecessary despite affecting volume flow rate transiently.
d. Neither (a) nor (b)
Wrong, since without glucose amount, π (and h) cannot be calculated.
Correct answer: a. The amount of glucose added.
Key Equation Derivation
π = CRT = (n/V)RT, with V = 1 L fixed initially. At equilibrium, ρgh = π, so h = (nRT)/(Vρg). Cross-sectional area A influences volume ΔV = A⋅(h/2) per arm but not final h, confirming only glucose amount (for n) is needed.
