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

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To lower the vapor pressure of 552 grams of water by 2 mmHg at 20 degrees Celsius, 83,000 grams of sucrose (C12H22O11) must be added. The calculation involves determining the mole fraction of sucrose needed to achieve this pressure reduction, which was found to be 0.11 moles. This corresponds to a total of 30.3 moles of water, leading to the conclusion that 245.18 moles of sucrose are required, equating to 83,000 grams when converted using the molar mass of sucrose (324.34 g/mol).

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

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