CBSE Class 12 · 2000 · Foreign · SolutionsShort Answer

An aqueous solution of glucose is made by dissolving 10 g of glucose (C

_{6}

_{12}

_{6}

) in 90 g of water at 303 K. If the vapour pressure of pure water at 303 K is 32.8 mm Hg, what would be the vapour pressure of the solution?

Answer: p ≈ 32.44 mm Hg

Solution

Moles of glucose:

n_{2} = \frac{10}{180} = 0.0556 mol

Moles of water:

n_{1} = \frac{90}{18} = 5 mol

Mole fraction of glucose:

x_{2} = \frac{0.0556}{5.0556} = 0.011

For a non-volatile solute,

p = p^{\circ} (1 - x_{2})

p = 32.8 \times (1 - 0.011)

p = 32.44 mm Hg

CBSE Class 12 Board · 2000 · Foreign · Subjective