Question

1. The vapor pressure of pure water at 26°Cis 25.21 mmHg. What is the vapor pressure of a solution which contains 179.38 g of iron (III)sulfate[Fe2(SO4)3]in 450.0 g of water?The molar mass of iron (III)sulfateand water are 399.88g/mol and 18.0 g/mol, respectively.

Answer 1

Answer: 26C is 25.0 torr

Explanation: The decrease in vapor pressure is proportional to the molality of the solution, so the vapor pressure of the solution at 26°C is 25.0 torr.

First, we need to find the number of moles of glucose and water present in the solution:

n(glucose) = 16.0 g / 180.0 g/mol = 0.089 mol

n(water) = 80.0 g / 18.0 g/mol = 4.44 mol

Next, we'll find the mole fraction of water in the solution:

X_water = n(water) / (n(glucose) + n(water)) = 4.44 / (0.089 + 4.44) = 0.989

Finally, we can find the vapor pressure of the solution using Raoult's law:

P_solution = X_water * P_purewater = 0.989 * 25.21 torr = 25.0 torr

So the vapor pressure of the solution at 26°C is 25.0 torr.