Answer:
[tex]23.093\ \text{torr}[/tex]
Explanation:
[tex]M_g[/tex] = Molar mass of glucose = 180.2 g/mol
[tex]M_w[/tex] = Molar mass of water = 18 g/mol
[tex]m_g[/tex] = Mass of glucose = 76.6 g
[tex]m_w[/tex] = Mass of water = [tex]250\times 1\ \text{g/mL}=250\text{g}[/tex]
[tex]P_0[/tex] = Vapor pressure of pure water at 25°C = 23.8 torr
The mole fraction of glucose is
[tex]x_g=\dfrac{\dfrac{m_g}{M_g}}{\dfrac{m_g}{M_g}+\dfrac{m_w}{M_w}}=\dfrac{\dfrac{76.6}{180.2}}{\dfrac{76.6}{180.2}+\dfrac{250}{18}}\\\Rightarrow x_g=0.0297[/tex]
Mole fraction of the solute would be
[tex]\dfrac{P_0-P}{P_0}=x_g\\\Rightarrow 0.0297=\dfrac{23.8-P}{23.8}\\\Rightarrow P=23.8-0.0297\times23.8\\\Rightarrow P=23.093\ \text{torr}[/tex]
The vapor pressure of the solution is [tex]23.093\ \text{torr}[/tex].
