Question

If a `"62.4 g/L"` solution of bovine insulin in an appropriate solvent achieves an osmotic pressure of `"0.305 atm"` at `"298 K"`, estimate the molar mass?

`a)` `"10000 g/mol"`
`b)` `"5000 g/mol"`
`c)` `"506887 g/mol"`
`d)` `"5069 g/mol"`

Answer 1

Formula Wt Insulin ~ 5000 g/mole*

Explanation:

Using the Osmotic Pressure Equation*

`Pi = MRT = ("moles"/"Vol"_L)*R*T`

=> `"moles" = (Pi*"Vol"_L)/("R"cdot"T") = "mass(g)"/"f.wt."`

=> `f.wt. = ("mass(g)"cdot"R"cdot"T")/(Picdot"Vol"_L)`

=> `f.wt. = (("62.4 g")("0.08206 L"cdot"atm"cdot"mol"^(-1)cdotK^(-1))("298 K"))/(("0.305 atm")("1.00 L")`

=> `f.wt. = "5000 g"/"mol"`

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Note: the method of mole weight analysis is not specified. That is, the 5808 mole wt. value is perhaps a weight average mole weight (mass fraction analysis), whereas the 5000 mole wt. calculation is based upon `Pi`, which is a colligative property and defines 'number average' mole weight values (size/chain length fraction analysis).

The calculation using the osmotic pressure equation is correctly done but is a colligative property dependent relationship which will typically give lower values than other methods defining weight average mole weights.