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If 79.5 g of glucose (molar mass = 180.2 g/mol) is dissolved in 38.4 g of water, what is the vapor pressure of the solution (in mmHg)?

Answer :

The vapor pressure of the solution is approximately 9.75 mmHg. To calculate vapor pressure of the solution, we can use Raoult's law:

Now,

First, find the moles of glucose (solute) and water (solvent):

Moles of Glucose = 79.5 g / 180.2 g/mol ≈ 0.441 mol

Moles of Water = 38.4 g / 18.015 g/mol ≈ 2.132 mol

Next, calculate the mole fractions:

Mole Fraction of Glucose = 0.441 / (0.441 + 2.132) ≈ 0.171

Mole Fraction of Water = 2.132 / (0.441 + 2.132) ≈ 0.829

The vapor pressure of pure water at given temperature is typically around 23.8 mmHg.

Raoult's law:-

Vapor Pressure of Solution = (0.829 * 23.8 mmHg) + (0.171 * 0 mmHg) ≈ 19.75 mmHg

Learn more about Raoult's law:-

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