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Step 1: Identify the Relevant Formula for Osmotic Pressure
The osmotic pressure, denoted by \Pi , can be related to the moles of solute using the equation:
\Pi = \frac{n}{V} \, R \, T,
where:
\Pi is the osmotic pressure (in bar),
n is the number of moles of the solute (protein, in this case),
V is the volume of the solution (in liters),
R is the gas constant (in \text{L bar K}^{-1} \text{mol}^{-1} ),
T is the temperature (in Kelvin).
Step 2: Convert All Given Data into Appropriate Units
Mass of protein, m = 0.63 \,\text{g}
Volume of solution, V = 300 \,\text{cm}^3 = 0.3 \,\text{L}
Osmotic pressure, \Pi = 1.29 \,\text{mbar} = 1.29 \times 10^{-3}\,\text{bar}
Temperature, T = 300 \,\text{K}
Gas constant, R = 0.083 \,\text{L bar K}^{-1}\text{mol}^{-1}
Step 3: Relate Moles to Molar Mass
The number of moles n of the protein is given by
n = \frac{m}{M},
where M is the molar mass of the protein (the unknown we need to find).
Substituting this expression into the osmotic pressure equation:
\Pi = \frac{\frac{m}{M}}{V} \, R \, T
\quad \Longrightarrow \quad
\Pi = \frac{m \, R \, T}{M \, V}.
Solving for M gives:
M = \frac{m \, R \, T}{\Pi \, V}.
Step 4: Plug in the Numerical Values
Using the values listed above:
\Pi = 1.29 \times 10^{-3}\,\text{bar}, \quad
V = 0.3\,\text{L}, \quad
R = 0.083\,\text{L bar K}^{-1}\text{mol}^{-1}, \quad
T = 300\,\text{K}, \quad
m = 0.63\,\text{g}.
Hence,
M
= \frac{0.63 \times 0.083 \times 300}{(1.29 \times 10^{-3}) \times 0.3}.
Step 5: Calculate the Molar Mass
• First, compute the numerator:
0.63 \times 0.083 \times 300 = 15.687.
• Then, the denominator:
(1.29 \times 10^{-3}) \times 0.3 = 3.87 \times 10^{-4}.
• Divide numerator by denominator:
M = \frac{15.687}{3.87 \times 10^{-4}} \approx 4.05 \times 10^4 \,\text{g mol}^{-1}.
Step 6: State the Final Result
Therefore, the molar mass of the protein is approximately
\boxed{4.05 \times 10^4 \,\text{g mol}^{-1}}.
