Question

A solution containing 28.45 mg of an unknown protein per 20.5 mL solution was found to have an osmotic pressure of 3.90 torr at 16 ∘ what is the molar mass?

Answer #1

6393.3g/mol

Explanation

π = MRT

where,

π = Osmotic pressure, 3.90torr=0.005148atm

M = molarity of solute

R = gas constant, 0.082057(L atm / K mol )

T = temperature, 16℃ = 289.15K

Therefore,

M = π / RT

= 0.005148atm/( 0.082057( L atm /mol K ) × 289.15K

= 2.17×10^-4M

2.17×10^-4M = 2.17×10^-4mol per Liter

No of mole in 20.5ml = (2.17×10^-4mol/1000ml)×20.5ml = 4.45×10^-6mol

Mass of 4.45×10^-6mol of protein = 28.45mg = 0.02845g

Molar mass = 0.02845g/4.45×10^-6mol = 6393.3g/mol

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