Question

Calculate the vapor pressure of a solution made by dissolving 11.1 g Ca(OH)2 in 1 102 g of water at 25 °C. Vapor pressure of pure water is 23.76 mm Hg at 25 °C.

Answer 1

Answer:

23.15 mmHg.

Explanation:

To solve this question you need to understand that part of Raoult's law in your Chemistry textbook. So, let us delve right into the solution of the question.

We are given parameters such as the mass of Ca(OH)2 to be = 11.1 grams, mass of solvent = 102 grams, the Vapor pressure of pure water= 23.76 mm Hg, temperature = 25°C and vapour pressure of the solution= ??.

The molar mass of Ca(OH)2= 74 g/mol.

The first thing to do is to find the number of moles of Ca(OH)2 and that of water from the formula below;

Mass/ molar mass = Number of moles.

===> Number of moles, n= 11.1/ 74.

Number of moles = 0.15 moles Ca(OH)2.

===> Number of moles, n= 102/ 18.

Number of moles, n= 5.67 moles of water.

Next, we add the two moles together to find the solvent mile fraction since vapour pressure is proportional to mole fraction.

Then;

0.15 + 5.67 = 5.82.

Therefore, 5.67/ 5.82= 0.97.

Hence the vapor pressure of a solution = 0.97 × 23.76.

vapor pressure of a solution = 23.15 mmHg.

Answer 2

Explanation:

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