Chp2_Osmotic

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

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Osmotic Pressure Example Ex: 2.52, p. 57 for statement of the problem. 1.5 g of enzyme in a 5.00 mL aqueous solution. = 0.213 atm at 25º C. Determine the molar mass of the enzyme. RT M c is our tool of choice. Note that i = 1 for a non-electrolyte (doesn’t break up into ions). L mol K K mol atm L atm RT M 3 1 1 10 71 . 8 298 0821 . 0 213 . 0 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 4.35e5 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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This note was uploaded on 03/29/2009 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.

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