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How many grams of propanol (C3H7OH, 60.10 g/mol) would be needed to make 750 mL of a solution with an osmotic pressure of 25 atm at 25°C? (R = 0.0821 L.atm/K.mol)
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Feb 6, 2021
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How many grams of propanol (C3H7OH, 60.10 g/mol) would be needed to make 750 mL of a solution with an osmotic pressure of 25 atm at 25°C? (R = 0.0821 L.atm/K.mol)
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ReggieB
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Answer:
= 46.06 grams
Step-by-step explanation:
P=cRT
25 = c (0.0821)(298)
concentration = c = 1.021 M
number of moles = (1.021)(0.75) = 0.7664 moles
mass = (0.7664)(60.10)=40.06 grams
Coli
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Feb 11, 2021
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