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Step-by-Step Solution
Step 1: Identify the known quantities
• Mass of protein, $w = 1.26 \text{ g}$
• Osmotic pressure, $\pi = 2.57 \times 10^{-3} \text{ bar}$
• Volume of solution, $V = 200 \text{ mL} = 0.2 \text{ L}$
• Temperature, $T = 300 \text{ K}$
• Gas constant, $R = 0.083 \text{ L bar mol}^{-1}\text{ K}^{-1}$
Step 2: Recall the formula relating osmotic pressure, molar mass, and other variables
The general relation for osmotic pressure is:
$$
\pi = \frac{n}{V} \, R T
$$
where:
• $\pi$ is the osmotic pressure
• $n$ is the number of moles of solute
• $V$ is the volume of solution (in liters)
• $R$ is the gas constant
• $T$ is the absolute temperature
We also know that the number of moles $n$ can be written as:
$$
n = \frac{w}{M}
$$
where $w$ is the mass of the solute (protein) and $M$ is its molar mass. Combining these, the formula for molar mass becomes:
$$
\pi V = \frac{w}{M} R T \quad \Longrightarrow \quad M = \frac{w \, R \, T}{\pi \, V}
$$
Step 3: Substitute the known values into the formula
$$
M = \frac{1.26 \times 0.083 \times 300}{2.57\times10^{-3}\times0.2}
$$
Step 4: Perform the calculation
First, calculate the numerator:
$$
1.26 \times 0.083 \times 300 = 1.26 \times (0.083 \times 300).
$$
Then compute the denominator:
$$
2.57 \times 10^{-3} \times 0.2.
$$
Finally, divide the numerator by the denominator to obtain $M$.
Step 5: Obtain the molar mass
After simplifying, we get:
$$
M \approx 61038 \text{ g mol}^{-1}.
$$
Therefore, the molar mass of the protein is approximately $61038 \text{ g mol}^{-1}$.
Answer
The molar mass of the protein is $61038 \text{ g mol}^{-1}$.
