Asked by Ruffles
What is the osmotic pressure of a 50% sucrose solution at a temperature of 37 degrees celcius with a molar mass of sucrose of 342.30 grams
Answers
Answered by
Jai
*repost*
Formula for osmotic pressure:
Π = MRT * i
where
m = molarity (mol solute / L solution)
R = universal gas constant = 0.0821 L-atm/mol-K
T = temperature (K)
i = van't Hoff factor (approximately equal to ionizable ions)
Note that sucrose is soluble in water, but does not form ions in water, thus i = 1.
I am not sure what that 50% sucrose solution means, so I assumed it's by mass, or 50 g sucrose over 100 g solution. Assuming density of solution is equal to that of water (which is 1 g/mL),
Π = MRT
Π = [(50/342)/((100/1)/1000)] * 0.0821 * (37 + 273)
Π = 37.21 atm
hope this helps :3
Formula for osmotic pressure:
Π = MRT * i
where
m = molarity (mol solute / L solution)
R = universal gas constant = 0.0821 L-atm/mol-K
T = temperature (K)
i = van't Hoff factor (approximately equal to ionizable ions)
Note that sucrose is soluble in water, but does not form ions in water, thus i = 1.
I am not sure what that 50% sucrose solution means, so I assumed it's by mass, or 50 g sucrose over 100 g solution. Assuming density of solution is equal to that of water (which is 1 g/mL),
Π = MRT
Π = [(50/342)/((100/1)/1000)] * 0.0821 * (37 + 273)
Π = 37.21 atm
hope this helps :3
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