`200" cm"^(2)` of an aqueous solution of a protein contains 1.26 g of the protein. The oxmotic pressure of such a solution at 300 K is found to be `2.

Question

`200" cm"^(2)` of an aqueous solution of a protein contains 1.26 g of the protein. The oxmotic pressure of such a solution at 300 K is found to be `2.

`200" cm"^(2)` of an aqueous solution of a protein contains 1.26 g of the protein. The oxmotic pressure of such a solution at 300 K is found to be `2.57xx10^(-3)` bar. Calculate the molar mass of the protein.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The various quantities known to us are as follows : `pi=2.57xx10^(-3)" bar".`
`V=200" cm"^(3)=0.200` litre
T = 300 K
`R=0.083" L bar mol"^(-1)"K"^(-1)`
Substituting these values in equation,
we get `M_(2)=(w_(2)RT)/(pi V)`
`M_(2)=(1.26" g"xx0.083" L bar K"^(-1)" mol"^(-1)xx300" K")/(2.57xx10^(-3)"bar"xx0.200" L")`
`=61.022" g mol"^(-1)`