A 48.4-g sample of glucose (a nondissociated, nonvolatile solute with the formula c6h12o6) is dissolved in 151.2 g of water. what is the vapor pressure of this solution at 100°c?

Vapor pressure of solution is 736mmHg

Explanation:

Vapor pressure of an ideal solution follows Raoult's law:

Where P is vapor pressure and X is mole fraction

Moles of glucose are:

48.4g × (1mol / 180.156g) = 0.2687mol glucose

Moles of water:

151.2g × (1mol / 18.1g) = 8.354 mol water

Thus, mole fraction of water (Solvent) is:

8.354 mol / (8.354 mol + 0.2687mol) = 0.9688

Vapor pressure of a solvent at boiling point is equal to atmospheric pressure (760mmHg). Replacing in Raoult's law:

Vapor pressure of solution is 736mmHg

the product of the pressure and volume is also constant