An aqueous solution of 10.00 g of an enzyme has a volume of 1.00 L at 35°C. The solution's osmotic pressure?
Full Question - An aqueous solution of 10.00 g of an enzyme has a volume of 1.00 L at 35°C. The solution's osmotic pressure at 35°C is found to be 0.713 torr. Calculate the molar mass of the enzyme.
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35°C-----> T = 308 K
V = 1 L
0.713 Torr----> Π = 0.0009382 atm
Π = MRT = (n/V) * RT
Number of moles = n = (Π * V) / (RT)
= (0.0009382 atm * 1 L) / (0.08206 L*atm*mol^-1*K^-1*308 K)
= 3.712 x 10^-5 mol
Molar Mass = Mass/Moles = Mass/n
= 10.00 g/3.712 x 10^-5 mol
= 269,397 g/mol <-----ANSWER