Question

Aqueous solutions introduced into the bloodstream by injection must have the same osmotic pressure as blood; that is, they must be "isotonic" with blood. At \(25^{\circ} \mathrm{C},\) the average osmotic pressure of blood is 7.7 atm. What is the molarity of an isotonic saline solution \(\left(\mathrm{NaCl}\right.\) in \(\left.\mathrm{H}_{2} \mathrm{O}\right)\) ? Recall that \(\mathrm{NaCl}\) is an electrolyte; assume complete conversion to \(\mathrm{Na}^{+}\) and \(C]^{-}\) inns

Short answer

Answer: The molarity of an isotonic saline solution at 25°C, with an osmotic pressure of 7.7 atm, is approximately 0.154 mol/L.

Step-by-step solution

Write down the van't Hoff equation

The van't Hoff equation relates osmotic pressure (Π), molarity (c), the ideal gas constant (R), and the temperature (T): Π = c * R * T * i Here, i is the van't Hoff factor, which accounts for the number of particles (ions) a solute produces in a solution. As the problem states that NaCl dissociates completely into Na⁺ and Cl⁻ ions, i = 2.

Convert values to corresponding units

To solve this equation, we need to express temperature in Kelvin and use the R value in L⋅atm/mol⋅K. Given the temperature is 25°C, we can find the corresponding temperature in Kelvin: T = 25°C + 273.15 = 298.15 K We will use the ideal gas constant, R, which is equal to 0.0821 L⋅atm⋅mol⁻¹⋅K⁻¹.

Substitute the data into the equation and solve for molarity

Given the average osmotic pressure of blood (Π) as 7.7 atm, we can now plug in the numbers into the van't Hoff equation: áπ = c * R * T * i Where Π = 7.7 atm R = 0.0821 L⋅atm⋅mol⁻¹⋅K⁻¹ T = 298.15 K i = 2 (for the dissociation of NaCl into Na⁺ and Cl⁻ ions) Solve for the molarity of the isotonic saline solution (c): ⟹ 7.7 = c * 0.0821 * 298.15 * 2 Divide both sides by 2 * 0.0821 * 298.15: c = 7.7 / (0.0821 * 298.15 * 2) = 0.154 mol/L

State the final answer

The molarity of an isotonic saline solution at 25°C, with an osmotic pressure of 7.7 atm, is approximately 0.154 mol/L.