Question 12 of 50intermediate🔧 ApplyNumerical2 marks

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.

Correct Answer

The osmotic pressure exerted by the solution is 30.96 Pa.

Given: Mass of polymer (w2w_2) = 1.0 g Molar mass (M2M_2) = 185,000 g mol⁻¹ Volume (VV) = 450 mL = 0.450 L = 0.450×10−3 m30.450 \times 10^{-3} \text{ m}^3 Temperature (TT) = 37°C = 310 K

Calculation: Using the formula for osmotic pressure:

Π=w2RTM2V\Pi = \frac{w_2 RT}{M_2 V}

Substituting values (using R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}):

Π=1.0 g×8.314 J K−1 mol−1×310 K185,000 g mol−1×0.450×10−3 m3\Pi = \frac{1.0 \text{ g} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 310 \text{ K}}{185,000 \text{ g mol}^{-1} \times 0.450 \times 10^{-3} \text{ m}^3} Π=2577.3483.25 Pa\Pi = \frac{2577.34}{83.25} \text{ Pa} Π=30.96 Pa\Pi = 30.96 \text{ Pa}
Exercise: Intext Questions | Q: 1.12 | (Page 23)
For More Understanding

Explanation

The textbook context provides the formula ΠV=w2RTM2\Pi V = \frac{w_2RT}{M_2}, which is derived from the proportionality of osmotic pressure to molarity (Π=CRT\Pi = CRT). This method is particularly useful for polymers and macromolecules because osmotic pressure measurements are precise even for dilute solutions at room temperature. To solve the numerical, one must convert the temperature to Kelvin and volume to SI units (cubic meters) to match the units of the gas constant RR (8.314 J K⁻¹ mol⁻¹), ensuring the final pressure is in Pascals.

Solution Steps

  1. Step 1: Identify the given values from the question: w2=1.0 gw_2 = 1.0 \text{ g}, M2=185,000 g mol−1M_2 = 185,000 \text{ g mol}^{-1}, V=450 mL=0.450×10−3 m3V = 450 \text{ mL} = 0.450 \times 10^{-3} \text{ m}^3, and T=37∘C=310 KT = 37^\circ\text{C} = 310 \text{ K}.

  2. Step 2: Write the formula for osmotic pressure provided in the context: Π=w2RTM2V\Pi = \frac{w_2RT}{M_2V}.

  3. Step 3: Substitute the values and the gas constant R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} into the equation.

  4. Step 4: Calculate the numerator (1.0×8.314×310=2577.341.0 \times 8.314 \times 310 = 2577.34) and the denominator (185,000×0.450×10−3=83.25185,000 \times 0.450 \times 10^{-3} = 83.25).

  5. Step 5: Divide the numerator by the denominator to find the osmotic pressure in Pascals (2577.34/83.25=30.96 Pa2577.34 / 83.25 = 30.96 \text{ Pa}).

PreviousNext