Question

100mg of a protein is dissolved in just enough water to make 10.0ml of solution. if this solution has an osmotic pressure of 13.3 mm hg at 25 0c, what is the molar mass of the protein? (r= 0.0821 l atm mol-1 k-1 and 760 mm hg = 1 atm)

Answer 1

Mass of Protein (W2)= 0.1
R= 0.082 L at atm/ K mol
T= 298 K
TT= 0.0175 atm
Volume of Solution = 0.01 L
Using, M2= W2RT/TTV
We get,
M2= 0.1 g x 0.82 L x 298 K/ 0.175 x 0.01 
M2= 13980.45
     =13,9 KDa

Answer 2

Answer: Molecular mass of protein is $1.4\times 10^{10}g/mol$

Explanation:

$\pi =CRT$

$\pi$ = osmotic pressure  = 13.3 mmHg = 0.0175 atm

C= concentration in Molarity

R= solution constant  = 0.0821 Latm/Kmol

T= temperature  =$25^0C=25+273=298K$

For the given solution: 100 mg or 100000 g of urea is dissolved to make 10 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$C_{protein}=\frac{100000\times 1000}{M\times 10}=\frac{10^7}M}$

$0.0175=\frac{10^7}{M}\times 0.0821\times 298}$

$M=1.4\times 10^{10}g/mol$