osmotic pressure

Question

A solution containing 0.5 g of a protein in 100 mL of solution has an osmotic pressure of 1.23 × 10⁻³ atm at 300 K. The molar mass of the protein is: (R = 0.082 L atm K⁻¹ mol⁻¹)

RankGuru · Accepted Answer

**Osmotic Pressure ($π$)** is defined by the formula:
$$\pi = \frac{n}{V}RT = \frac{w R T}{M V}$$
- **Given**:
  - $\pi = 1.23 	imes 10^{-3} 	ext{ atm}$
  - $w = 0.5 	ext{ g}$, $V = 100 	ext{ mL} = 0.1 	ext{ L}$
  - $R = 0.082 	ext{ L atm K}^{-1} 	ext{ mol}^{-1}$, $T = 300 	ext{ K}$
- **Calculation**:
$$M = \frac{w R T}{\pi V} = \frac{0.5 	imes 0.082 	imes 300}{1.23 	imes 10^{-3} 	imes 0.1}$$
$$M = \frac{12.3}{1.23 	imes 10^{-4}} = 10^5 = 100,000 	ext{ g/mol}$$.

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More osmotic pressure Practice Questions

Osmotic pressure (π) of a dilute solution relates to concentration via: