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wariber [46]
3 years ago
6

The vapor pressure of pure water at 250C is 23.77 torr. What is the vapor pressure of water above a solution that is 1.500 m glu

cose, C6H12O6?
Chemistry
1 answer:
Olegator [25]3 years ago
7 0

Answer:

Vapor pressure of water = 23.14torr

Explanation:

When you made a solution, vapor pressure decreases following Raoult's law:

P_{solution} = X_{solvent} P_{solvent}

<em>Where P is vapor pressure and X mole fraction</em>

As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).

1000g of H₂O are, in moles (Molar mass: 18.02g/mol):

1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.

As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:

X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737

Replacing in Raoult's law, pressure of water above the solution is:

P_{solution} = X_{solvent} P_{solvent}

P_{solution} = 0.9737*23.77torr

<h3>Vapor pressure of water = 23.14torr</h3>
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Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

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Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

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For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

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