A solution is prepared by dissolving 60.0 g of sucrose, C12H22O11, in 250. g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?

23.46 mmHg is the vapor pressure for the solution

Explanation:

To solve this problem we need to apply a colligative property, which is the lowering vapor pressure.

The formula for this is: P°- P' = P° . Xm

where P' is vapor pressure for solution and P°, vapor pressure for pure solvent.

Let's determine the Xm (mole fraction for solute)

We calculate the moles of the solute and the solvent and we sum each other:

Moles of solute: 60 g /342 g/mol = 0.175 moles of sucrose

Moles of solvent: 250 g / 18 g/mol = 13.8 moles of water

Total moles: 13.8 moles + 0.175 moles = 13.975 moles

Xm for solute: 0.175 moles / 13.975 moles = 0.0125

Let's replace data in the formula: 23.76 mmHg - P' = 23.76 mmHg . 0.0125

P' = - (23.76 mmHg . 0.0125 - 23.76 mmHg) → 23.46 mmHg

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.