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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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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)

User ReggieB
by
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1 Answer

2 votes
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
User Coli
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