Answer:
![23.093\ \text{torr}](https://img.qammunity.org/2021/formulas/chemistry/college/bl11zqndz8s2o5xp9bk9aii52ogul6jedx.png)
Step-by-step explanation:
= Molar mass of glucose = 180.2 g/mol
= Molar mass of water = 18 g/mol
= Mass of glucose = 76.6 g
= Mass of water =
![250* 1\ \text{g/mL}=250\text{g}](https://img.qammunity.org/2021/formulas/chemistry/college/jmo7qaihlpfzryjhi44ws9wmw877m20dwy.png)
= 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](https://img.qammunity.org/2021/formulas/chemistry/college/6jio9t69bshpn2dloodlv5pu31wongp0ao.png)
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}](https://img.qammunity.org/2021/formulas/chemistry/college/yyhjtcrjehd9uszc23cptcf9k885qj28pd.png)
The vapor pressure of the solution is
.