Question

the osmotic pressure ofa solution containing 5.87 mg of an unknown protein per 10ml of solution is 2.45 torr at 25 degree celsius. Find the molar mass of the unknown protein.

Answer 1

Answer: The molar mass of the unknown protein is $4.45\times 10^9g/mol$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = 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 = $62.364\text{ L.torr }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.45torr=1\times \frac{5870\times 1000}{\text{Molar mass of protein}\times 10}\times 62.364\text{ L.torr}mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}=4.45\times 10^9g/mol$

Hence, the molar mass of the protein is $4.45\times 10^9g/mol$

Answer 2

The molar mass of protein will be "4.45×10⁹ g/mol".

Given values:

• Osmotic pressure, $\pi = 2.45 \ torr$
• Van't hoff factor, $i = 1$
• Mass, $m = 5.87 \ mg \ or \ 5870 \ g$
• Volume, $V = 10 \ mL$
• Gas constant, $R = 62.364 \ L.torr \ mol^{-1} K^{-1}$
• Temperature, $T = 25^{\circ} C$

                                    $= 273+25$

                                    $= 298 \ K$

We know that,

→                 $\pi = iMRT$

or,

→                 $\pi = i\times \frac{Mass \ of \ solute\times 1000}{Molar \ mass \ of \ solute\times Volume}\times RT$

By substituting the values, we get

The Molar Mass:

→             $2.45=1\times \frac{5870\times 1000}{Molar \ mass \ of \ protein\times 10}\times 62.364\times 298$

  $Molar \ mass = 4.45\times 10^9 \ g/mol$      

Thus the response above is right.

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