The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL⁻¹ is _____________.
(Molar mass of glucose = 180 g mol⁻¹, R = 0.083 L bar K⁻¹ mol⁻¹) (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Isotonic means same osmotic pressure, so find the molarity from blood and turn it into grams.

1. Isotonic condition. An isotonic glucose solution has the same osmotic pressure as blood, 7.47 bar at 300 K. Glucose does not dissociate in water, so i = 1.

   π = CRT ⇒ C = π/RT

2. Find molarity. RT = 0.083 × 300 = 24.9 L bar mol⁻¹. So C = 7.47/24.9 = 0.3 mol L⁻¹.
3. Convert to grams. Multiply by the molar mass: 0.3 mol L⁻¹ × 180 g mol⁻¹ = 54 g L⁻¹.
4. So. The glucose solution must contain 54 g L⁻¹.

The common slip: 0.3. Stopping at the molarity gives 0.3, which is in mol L⁻¹. Multiply by 180 to get g L⁻¹.

The idea to keep. Isotonic solutions share the same osmotic pressure. Use π = iCRT to get molarity, then multiply by molar mass whenever the answer is asked in g L⁻¹.

Answer from the NTA final answer key. Solution by PrepWiser.