The vapour pressure of pure water at 25°c is 18 mm Hg . When a nonelectrolyte is dissolved, its vapour pressure becomes 17.32 mm Hg . What is the relative lowering of vapour pressure of the solution?

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The vapour pressure of pure water at 25°c is 18 mm Hg . When a nonelectrolyte is dissolved, its vapour pressure becomes 17.32 mm Hg . What is the relative lowering of vapour pressure of the solution?

(vii) The vapour pressure of pure water at $ 25^{\circ} C $ is $ 18 mm Hg $. When a nonelectrolyte is dissolved, its vapour pressure becomes $ 17.32 mm Hg $. What is the relative lowering of vapour pressure of the solution?

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Salim Ahmed Kawsar · Answer 1 · Feb 05, 2023 15:29:48

We have given,

The vapour pressure of pure water at 25°C = 18mm Hg.

After addition of non-electrolyte, the vapour pressure of water = 17.32 mm Hg.

Then,

lowering in vapour pressure = (18 - 17.32) mm Hg. = 0.68 mm Hg.

∴ relative lowering in vapour pressure $=\frac{0.68\,mmHg}{18.0\,mmHg}=0.038$

The vapour pressure of pure water at 25°c is 18 mm Hg . When a nonelectrolyte is dissolved, its vapour pressure becomes 17.32 mm Hg . What is the relative lowering of vapour pressure of the solution?

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