Question 6.12: Calculate the osmotic pressure of a 5.0 × 10^–2 M solution o...
Calculate the osmotic pressure of a 5.0 \times 10^{–2} M solution of NaCl at 25°C (298 K).
Step 1. Using our definition of osmotic pressure, π:
π = MRT
Step 2. M should be represented as osmolarity as we have shown in Example 6.11
M = 5.0 \times 10^{-2} \frac{\cancel{mol NaCl}}{L} \times \frac{2 mol particles}{1 \cancel{mol NaCl}} = 1.0 \times 10^{-1} \frac{mol particles}{L}
Step 3. Substituting in our osmotic pressure expression:
\pi = 1.0 \times 10^{-1} \frac{\cancel{mol particles}}{\cancel{L} } \times 0.0821 \frac{\cancel{L} – atm}{\cancel{K} – \cancel{mol}}\times 298 \cancel{K} \\ \; \; \; = 2.4 atmRelated Answered Questions
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