Question

In a laboratory experiment, students synthesized a new compound and found that when 11.56 grams of the compound were dissolved to make 280.0 mL of a ethanol solution, the osmotic pressure generated was 3.50 atm at 298 K. The compound was also found to be nonvolatile and a nonelectrolyte. What is the molecular weight they determined for this compound? g/mol

Answer 1

Step 1

The osmotic pressure is calculated as follows:

`\begin{gathered} \pi\text{ = C x R x T} \\ C\text{ = molarity = }\frac{moles\text{ }ofsolute}{Volume\text{ of solution \lparen L\rparen}} \\ Moles\text{ of solute = }\frac{mass\text{ of solute}}{M} \end{gathered}`

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Step 2

Information provided:

The mass of solute = 11.56 g

Volume of solution = V = 280.0 mL

(1 L = 1000 mL => V = 0.280 L)

π = 3.50 atm

Absolute temperature = T = 298 K

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Information needed:

R = 0.082 atm L/mol K

M = the molar mass = unknown

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Step 3

Procedure:

From step 1:

π = C x R x T

π = (Moles of solute/V) x R x T

π = [mass of solute/(M x V)] x R x T

So, M:

M = [Mass of solute/(π x V)] x R x T

M = [11.56 g/(3.50 atm x 0.280 L)] x 0.082 atm L/mol K x 298 K

M = 288.2 g/mol

Answer: M = 288.2 g/mol