An aqueous solution of volume 300 cm 3 contains 0 . 63 g of protein. The osmotic pressure of the solution at 300 K is 1 . 29 mbar . The molar mass of the protein is gmol - 1 . Given : R = 0 . 083 L ba

Question

An aqueous solution of volume 300 cm 3 contains 0 . 63 g of protein. The osmotic pressure of the solution at 300 K is 1 . 29 mbar . The molar mass of the protein is gmol - 1 . Given : R = 0 . 083 L bar K - 1 mol - 1

TestHub · Accepted Answer

The osmotic pressure of a non-electrolytic solution can be calculated as follows,

$π = CRT$

$π$ = osmotic pressure

$C$ = molarity $= \frac{Number of moles}{Volume of solution in L}$

$(1.29 \times 10^{- 3}) = \frac{0.63}{(MW) (0.3)} \times (0.083) \times 300$

$MW = \frac{0.63 \times 0.083 \times 300}{(0.3) \times (1.29 \times 10^{- 3})}$

$= 40534.88$

$\approx 40535$ (Nearest integer)

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