Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

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Solution

Answer:
Vapour pressure of water

p_{A}^{0} = 17.535 mm of Hg

Mass of glucose W₂ = 25 g
mass of water w₁ = 450g
we know that
mass of glucose = 180gmol^-1
molar mass of glucose =25/180g mol^-1
mass of water = 18 g
molar mass of water M₂ = 450/18 g mol^-1apply equation we get,

\frac{p_{A}^{0} - p_{A}}{p_{A}^{0}} = x_{A}

\frac{17.535 - p_{A}}{17.535} = \frac{25 / 180}{450 / 18}

\frac{17.535 - p_{A}}{17.535} = \frac{25}{180} \times \frac{18}{450} = \frac{1}{180}

180 (17.535 - p_{A}) = 17.535

3156.30 - 180 p_{A} = 17.535

3138.765 = 180 p_{A}

p_{A} = \frac{3138.765}{180} = 17.44 mm Hg .

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Solutions

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Some More Questions From Solutions Chapter

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 37⁰C.

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For Daily Free Study Material Join wiredfaculty

Solutions

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Some More Questions From Solutions Chapter

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 370C.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 37⁰C.