Calculate the osmotic pressure at 25°C and the freezing point of a 1.8% aqueous solution of glucose (C$_6$H$_{12}$O$_6$). Assume ideal behaviour of the solution. Take density to be 1 g mL$^{-1}$ and K$_f$ for water to be 1.86 K kg mol$^{-1}$. (R = 0.082 L atm mol$^{-1}$ K$^{-1}$)

CBSE Class 12 · 2005 · Delhi Compartment · SolutionsShort Answer

Calculate the osmotic pressure at 25°C and the freezing point of a 1.8% aqueous solution of glucose (C

_{6}

_{12}

_{6}

). Assume ideal behaviour of the solution. Take density to be 1 g mL

^{- 1}

and K

_{f}

for water to be 1.86 K kg mol

^{- 1}

. (R = 0.082 L atm mol

^{- 1}

^{- 1}

)

Answer: π ≈ 2.44 atm; T_f ≈ −0.19°C

1.8% solution: 1.8 g glucose in 100 mL (= 100 g) of solution.
Moles of glucose =

1.8/180 = 0.01

mol.
Osmotic pressure (V = 0.1 L):

π = \frac{n}{V} R T = \frac{0.01}{0.1} \times 0.082 \times 298

π = 2.44 atm

Mass of water =

100 - 1.8 = 98.2

g, so molality:

m = \frac{0.01}{0.0982} = 0.1018 mol kg^{- 1}

Depression:

Δ T_{f} = 1.86 \times 0.1018 = 0.19 K

T_{f} = - 0.19^{\circ} C

CBSE Class 12 Board · 2005 · Delhi Compartment · Subjective