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Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

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It is given that:
Volume of water, \(\mathrm{V}=450 \mathrm{~mL}=0.45 \mathrm{~L}\)
Temperature, \(\mathrm{T}=(37+273) \mathrm{K}=310 \mathrm{~K}\)
Number of moles of the polymer, \(n=\frac{1}{185000} \mathrm{~mol}\) We know that:
Osmotic pressure, \(\pi=\frac{n}{V} \mathrm{R} T\)
$$
\begin{aligned}
&=\frac{1}{185000} \mathrm{~mol} \times \frac{1}{0.45 \mathrm{~L}} \times 8.314 \times 10^{3} \mathrm{~Pa} \mathrm{~L} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 310 \mathrm{~K} \\
&=30.98 \mathrm{~Pa} \\
&=31 \mathrm{~Pa} \text { (approximately) }
\end{aligned}
$$
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