Calculate Protein Molar Mass Using Osmotic Pressure | Homework Example"

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SUMMARY

The discussion focuses on calculating the molar mass of a protein using osmotic pressure, specifically with a solution containing 35.0 g of protein in 1.0 L of water, resulting in an osmotic pressure of 10.0 mm Hg at 25°C. The osmotic pressure formula, π = MRT, is utilized, converting pressure to atm and calculating molarity as 5.37 x 10^-4 moles/liter. The molar mass is then derived by dividing the mass of the protein by the calculated moles, confirming the method's validity for varying quantities and volumes of protein solutions.

PREREQUISITES
  • Understanding of osmotic pressure and its formula (π = MRT)
  • Knowledge of unit conversions, particularly between mm Hg and atm
  • Familiarity with molarity calculations
  • Basic principles of stoichiometry related to mass and moles
NEXT STEPS
  • Study the impact of temperature on osmotic pressure calculations
  • Learn about the van 't Hoff factor and its effect on osmotic pressure
  • Explore advanced molarity calculations in multi-component solutions
  • Investigate real-world applications of osmotic pressure in biochemistry
USEFUL FOR

Students in chemistry or biochemistry courses, educators teaching osmotic pressure concepts, and researchers involved in protein analysis and molar mass determination.

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Homework Statement



A solution is prepared by dissolving 35.0 g of a protein in enough water to make up 1.0 L in volume. If the osmotic pressure of the solution is found to be 10.0 mm Hg at 25 C, the molar mass of the protein will be closest to:

Homework Equations



Osmotic pressure = MRT

The Attempt at a Solution



Pressure is expressed in atm, so 10/760 = 0.013 atm = M(0.0821)(25 + 273 K)

The molarity = 5.37 * 10^-4 moles/liter.

Now, we are given that there are 35 grams of protein per 1 liter of water in the problem.

(35g/L)/(5.37 * 10^-4 moles/liter) = molar mass of the protein, since molar mass is mass/moles.

The units being divided are (grams/liter) and (moles/liter). This equals:

Grams/liter * (liters / moles)

The liters cancel out, and we get grams / mole.

Question:

1) I'm fairly confident my work is correct and my answer is similarly correct.

2) However, if this problem were to be written, say, there are 70 grams of protein in enough water to make up 2.0 L in volume, I would have to do this, right?

(70 grams / 2 liters)/molarity of the solution? Because grams/liters divided by moles/liter is equivalent to grams/liters * liters/moles and the liters cancel out and we get grams/mole.

And if the problem were to state 70 grams / 2 liters I would be erroronus to perform 70 grams / molarity, since there are 2 LITERS of water, right? Doing this would be assuming there were only 1 liter of water.
 
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Typically if units work, result is OK. That's what the units are for.
 
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