Question

Determine the vapor pressure of a solution at 25°C that contains 76.6 g of glucose (C6H12O6 ) in 250.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr

Answer 1

`23.093\ \text{torr}`

Step-by-step explanation:

`M_g` = Molar mass of glucose = 180.2 g/mol

`M_w` = Molar mass of water = 18 g/mol

`m_g` = Mass of glucose = 76.6 g

`m_w` = Mass of water =
`250* 1\ \text{g/mL}=250\text{g}`

`P_0` = Vapor pressure of pure water at 25°C = 23.8 torr

The mole fraction of glucose is

`x_g=((m_g)/(M_g))/((m_g)/(M_g)+(m_w)/(M_w))=((76.6)/(180.2))/((76.6)/(180.2)+(250)/(18))\\\Rightarrow x_g=0.0297`

Mole fraction of the solute would be

`(P_0-P)/(P_0)=x_g\\\Rightarrow 0.0297=(23.8-P)/(23.8)\\\Rightarrow P=23.8-0.0297*23.8\\\Rightarrow P=23.093\ \text{torr}`

The vapor pressure of the solution is
`23.093\ \text{torr}`.