Question

a student dissolves 20.0 g of glucose into 511 mL of water at 25 C the vapor pressure of pure water at 25 C is 3.13

Answer 1

3.12 × 10⁻² atm

Step-by-step explanation:

A student dissolves 20.0 g of glucose into 511 mL of water. At 25 °C, the vapor pressure of pure water at 25 C is 3.13 × 10⁻² atm. I think the question is: "What is the vapor pressure of the solution?"

According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent present.

`P_(solution)=P\°_(solvent)X_(solvent)`

The molar mass of glucose is 180.16 g/mol. The moles corresponding to 20.0 g of glucose are:

20.0 g × (1 mol/180.16 g) = 0.111 mol

The density of water at 25°C is 0.997 g/mL. The mass corresponding to 511 mL of water is:

511 mL × (0.997 g/mL) = 509 g

The molar mass of water is 18.02 g/mol. The moles corresponding to 509 g of water are:

509 g × (1 mol/18.02 g) = 28.2 mol

The total number of moles is 0.111 mol + 28.2 mol = 28.3 mol

The mole fraction of water is:

`X_(solvent)=(28.2mol)/(28.3mol) =0.996`

The vapor pressure of a solvent above the solution is:

`P_(solution)=3.13 * 10^(-2) atm * 0.996 = 3.12 * 10^(-2) atm`