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Step-by-Step Solution
Step 1: Identify the known parameters
• Mass of solute, w = 2.5\,\mathrm{g}
• Volume of solution, V = 250.0\,\mathrm{mL} = 0.25\,\mathrm{L}
• Osmotic pressure, \pi = 400\,\mathrm{Pa}
• Temperature, T = 27^\circ\mathrm{C} = 300\,\mathrm{K}
• Gas constant, R = 0.083\,\mathrm{L\,bar\,K^{-1}\,mol^{-1}} (given)
Step 2: Convert osmotic pressure to bar
1 bar = 10^5\,\mathrm{Pa} . Therefore,
\[
\pi = 400\,\mathrm{Pa}
= \frac{400}{10^5}\,\mathrm{bar}
= 0.004\,\mathrm{bar}.
\]
Step 3: Write the expression for osmotic pressure
Osmotic pressure ( \pi ) is given by the formula
\[
\pi = \frac{n}{V}\,R\,T,
\]
where n is the number of moles of solute, V is the volume in liters, R is the gas constant, and T is the temperature in kelvin. We can also express n as \frac{w}{M} , where w is the mass of the solute and M is its molar mass. Thus,
\[
\pi = \frac{\frac{w}{M}}{V} \, R \, T \; = \; \frac{w}{M\,V}\,R\,T.
\]
Step 4: Substitute the values and solve for the molar mass
\[
0.004
= \frac{2.5}{M \times 0.25} \,\times\, (0.083) \,\times\, 300.
\]
First, simplify the multiplication on the right side:
\[
(0.083) \times 300 = 24.9,
\quad
\frac{1}{0.25} = 4.
\]
So the equation becomes
\[
0.004 = \frac{2.5}{M} \times 4 \times 24.9
= \frac{2.5 \times 99.6}{M}
= \frac{249}{M}.
\]
Rearranging to solve for M :
\[
M = \frac{249}{0.004} = 62250\,\mathrm{g\,mol^{-1}}.
\]
Step 5: State the final answer
Hence, the molar mass of the solute (to the nearest integer) is
\[
\boxed{62250\,\mathrm{g\,mol^{-1}}}.
\]
