The molar mass of a protein if a 0. 30 l of solution containing 0. 45 g of the protein has an osmotic pressure of 0. 80 torr at 25∘c is 0.348 g/mol
Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.80 - ( given ). This is only one part of the information we are given / can conclude in this case ....
i = van’t Hoff factor = 1 for a protein molecule,
R = gas constant = 62.36 L torr / K-mol,
T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K
( Known initially ) ∏ = osmotic pressure = 0.80 torr
besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.
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Substitute derived / known values to solve for M ( moles / liter ) -
∏ = iMRT
⇒ 0.80 = ( 1 )( M )( 62.36 )( 298 )
⇒ 0.80 = M( 18583.28 )
⇒ M = 0.80 / 18583.28 ≈ 4.3049 ....
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We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -
M = mol / l
⇒ 4.3049 = 0.045 / 25 mL ( 0.030 L ),
0.045 / 0.030 = 1.5 g / L
⇒ 1.5 g = 4.3049 moles,
molar mass = 1.5 g = 4.3049 moles = 0.348 g / mol
Thus the concluded that the molar mass of protein is 0.348 g / mol.
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