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This figure shows three scenarios relating to red blood cell membranes. In a, H subscript 2 O has two arrows drawn from it pointing into a red disk. Beneath it in a circle are eleven similar disks with a bulging appearance, one of which appears to have burst with blue liquid erupting from it. In b, the image is similar except that rather than having two arrows pointing into the red disk, one points in and a second points out toward the H subscript 2 O. In the circle beneath, twelve of the red disks are present. In c, both arrows are drawn from a red shriveled disk toward the H subscript 2 O. In the circle below, twelve shriveled disks are shown.
Red blood cell membranes are water permeable and will (a) swell and possibly rupture in a hypotonic solution; (b) maintain normal volume and shape in an isotonic solution; and (c) shrivel and possibly die in a hypertonic solution. (credit a/b/c: modifications of work by “LadyofHats”/Wikimedia commons)

Determination of molar masses

Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.

Determination of a molar mass from a freezing point depression

A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?


We can solve this problem using the following steps.

This is diagram with five boxes oriented horizontally and linked together with arrows numbered 1 to 4 pointing from each box in succession to the next one to the right. The first box is labeled, “Freezing point of solution.” Arrow 1 points from this box to a second box labeled, “delta T subscript f.” Arrow 2 points from this box to to a third box labeled “Molal concentration of compound.” Arrow labeled 3 points from this box to a fourth box labeled, “Moles of compound in sample.” Arrow 4 points to a fifth box labeled, “Molar mass of compound.”
  1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene ( [link] ).
    Δ T f = 5.5 ° C 2.32 ° C = 3.2 ° C
  2. Determine the molal concentration from K f , the freezing point depression constant for benzene ( [link] ), and Δ T f .
    Δ T f = K f m m = Δ T f K f = 3.2 ° C 5.12 ° C m −1 = 0.63 m
  3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
    Moles of solute = 0.62 mol solute 1.00 kg solvent × 0.0550 kg solvent = 0.035 mol
  4. Determine the molar mass from the mass of the solute and the number of moles in that mass.
    Molar mass = 4.00 g 0.034 mol = 1.2 × 10 2 g/mol

Check your learning

A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?


1.8 × 10 2 g/mol

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Determination of a molar mass from osmotic pressure

A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?


Here is one set of steps that can be used to solve the problem:

This is a diagram with four boxes oriented horizontally and linked together with arrows numbered 1 to 3 pointing from each box in succession to the next one to the right. The first box is labeled, “Osmotic pressure.” Arrow 1 points from this box to a second box labeled, “Molar concentration.” Arrow 2 points from this box to to a third box labeled, “Moles of hemoglobin in sample.” Arrow labeled 3 points from this box to a fourth box labeled, “Molar mass of hemoglobin.”
  1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.
    Π = 5.9 torr × 1 atm 760 torr = 7.8 × 10 −3 atm Π = MRT M = Π R T = 7.8 × 10 −3 atm ( 0.08206 L atm/mol K ) ( 295 K ) = 3.2 × 10 −4 M
  2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
    moles of hemoglobin = 3.2 × 10 −4 mol 1 L solution × 0.500 L solution = 1.6 × 10 −4 mol
  3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
    molar mass = 10.0 g 1.6 × 10 −4 mol = 6.2 × 10 4 g/mol

Check your learning

What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?


2.7 × 10 4 g/mol

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Colligative properties of electrolytes

As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.

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