Final answer:
To achieve an osmotic pressure of 0.75 atm in 2.5 L of water with CaCl₂ at 300 K, approximately 0.8372 grams must be dissolved, based on the van't Hoff equation and the assumption of ideal dissociation behavior.
Step-by-step explanation:
To calculate the mass of CaCl₂ that needs to be dissolved in 2.5 L of water to achieve an osmotic pressure of 0.75 atm at 300 K, we can use the van't Hoff equation for osmotic pressure: Π = iMRT, where Π is the osmotic pressure, i is the van 't Hoff factor, M is the molarity (mol/L),
R is the gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature in Kelvin. Calcium chloride (CaCl₂) dissociates into one calcium ion (Ca²⁺) and two chloride ions (Cl⁻), giving it a van 't Hoff factor (i) of 3 (ideal). Solving for M (M = Π / (iRT)), we get M = 0.75 / (3⋅0.0821⋅300) = 0.003015 mol/L. The molar mass of CaCl₂ is approximately 110.98 g/mol, so the total mass required for a 2.5 L solution is 2.5 L ⋅ 0.003015 mol/L ⋅ 110.98 g/mol = 0.8372 grams.