0.500 grams of a non-dissociating protein are dissolved in enough water to make
ID: 993358 • Letter: 0
Question
0.500 grams of a non-dissociating protein are dissolved in enough water to make 50.0 mL of solution. The osmotic pressure of this solution is measured to be 5.40 torr at 25°C. Determine the molar mass of this protein. (Note: 760 torr = 1 atm)
The equation for osmotic pressure is:
V = nRT
where = osmotic pressure in atm, V = volume of solution in L, n = number of moles of solute, R = 0.0821 Latm/molK, and T = temperature in Kelvin.
Using this equation, you can plug in all of the data given in this problem (be sure to convert everything to the correct units!) and then solve for n = # of moles of solute.
Once you have the moles of solute that are present, you can calculate the molar mass of the solute (the protein in this case) by dividing the mass (0.500 g) by the number of moles.
Answer options:
a) 34.4 g/mol
b) 34,400 g/mol
c) 45,300 g/mol
d) 68,900 g/mol
Explanation / Answer
V = nRT
V = ( m / M ) RT
M = m x R x T / V
M = 0.500 x 0.0821 x 298 / (5.40/760) 0.050
M = 34433 g/mol
Therefore answer is option ( b) 34,400 g/mol
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