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Chp2_Osmotic

# Chp2_Osmotic - In that 5.00 mL therefore are enzyme mol...

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Osmotic Pressure Example Ex: 2.52, p. 61 for statement of the problem. 1.5 g of enzyme in a 5.00 mL aqueous solution. Π = 0.213 atm at 25º C. Determine the molar mass of the enzyme. RT M c = Π is our tool of choice. Note that i = 1 for a non-electrolyte (doesn’t break up into ions). ( 29 L mol K K mol atm L atm RT M 3 1 1 10 71 . 8 298 0821 . 0 213 . 0 - - - × = × = Π = This is the concentration of a sample whose actual “size” or volume is 5.00 mL.
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Unformatted text preview: In that 5.00 mL therefore are: enzyme mol mmol mol mL mmol mL 5 3 10 35 . 4 1000 1 10 71 . 8 00 . 5--× = × × × . That is to say, that 4.35e–5 mol of enzyme has a mass of 1.50 g ( vide supra ). The molar mass is nothing other than the ratio of the mass to number of mols, viz. mol g mol g M m / 10 45 . 3 10 35 . 4 50 . 1 4 5 × = × =-...
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