Answer: The molar mass of the unknown protein is [tex]4.45\times 10^9g/mol[/tex]
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
Or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 2.45 torr
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (protein) = 5.87 mg = 5870 g (Conversion factor: 1 g = 1000 mg)
Volume of solution = 10 mL
R = Gas constant = [tex]62.364\text{ L.torr }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[273+25]=298K[/tex]
Putting values in above equation, we get:
[tex]2.45torr=1\times \frac{5870\times 1000}{\text{Molar mass of protein}\times 10}\times 62.364\text{ L.torr}mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}=4.45\times 10^9g/mol[/tex]
Hence, the molar mass of the protein is [tex]4.45\times 10^9g/mol[/tex]
Given values:
[tex]= 273+25[/tex]
[tex]= 298 \ K[/tex]
We know that,
→ [tex]\pi = iMRT[/tex]
or,
→ [tex]\pi = i\times \frac{Mass \ of \ solute\times 1000}{Molar \ mass \ of \ solute\times Volume}\times RT[/tex]
By substituting the values, we get
The Molar Mass:
→ [tex]2.45=1\times \frac{5870\times 1000}{Molar \ mass \ of \ protein\times 10}\times 62.364\times 298[/tex]
[tex]Molar \ mass = 4.45\times 10^9 \ g/mol[/tex]
Thus the response above is right.
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