How Much Sucrose Is Needed to Lower Water's Vapor Pressure by 2 mmHg?

[sw3etazngyrl]

How many grams of sucrose (C12H22O11) must be added to 552 g of water to give a solution with a vapor pressure 2.0 mmHg less than that of pure water at 20 degrees C? (The vapor pressure of water at 20 degrees C is 17.5 mmHg.)

I tried it out, and this is my work.
change in P=mole fraction*P in degrees
2.0mmHg=X(17.5mmHg)
X=0.11 moles

552 g H2O * 1 mole/18.02g = 30.3 mol

0.11(30.3 + x)=30.3
3.33+0.11x=30.3
x=245.18 mol

245.18 mol sucrose * 324.34g/1 mol=83000

 

[Gokul43201]

Hello sw3etazngyrl, we need to first see your effort. https://www.physicsforums.com/showthread.php?t=94384

 

[Blueshift5]

.4 g or 83 kg

Based on the given information, the solution would require approximately 83 kilograms of sucrose (C12H22O11) to be added to 552 grams of water to achieve a vapor pressure 2.0 mmHg less than that of pure water at 20 degrees C. This calculation takes into account the colligative properties of the solution, specifically the decrease in vapor pressure caused by the addition of a solute. It is important for scientists to understand colligative properties in order to accurately predict and control the behavior of solutions, as well as to develop various applications in fields such as chemistry, biology, and engineering. Further experimentation and analysis may be necessary to determine the exact amount of sucrose needed for the desired change in vapor pressure, as other factors such as temperature and solute-solvent interactions may also play a role.