What is the osmolarity of a 3.3%(w/v) Na₃PO₄ solution? A) 0.20 osmol B) 0.40 osmol C) 0.80 osmol D) 1.6 osmol

Question

Quizplus · Accepted Answer

To calculate osmolarity, we need to know the molar concentration of the solution. We can start by converting the percentage to grams per liter: 3.3 g/100 mL = 33.0 g/L Next, we need to calculate the molarity of Na₃PO₄. The molecular weight of Na₃PO₄ is 142 g/mol (22.99 for Na, 32.06 for ₃, 15.99 for O, and 35.45 for Cl). 33.0 g/L / 142 g/mol = 0.232 mol/L Finally, we convert the molarity to osmolarity by multiplying by the number of particles the molecule dissociates into in solution. Na₃PO₄ dissociates into 2 ions (Na+ and Cl-), so: 0.232 mol/L x 2 = 0.464 osmol/L However, the question asks for the osmolarity of the solution, so we need to convert from osmol/L to osmol: 0.464 osmol/L * 1 L/1000 mL * 1000 mL = 0.464 osmol Therefore, the osmolarity of a 3.3%(w/v) Na₃PO₄ solution is C) 0.80 osmol.

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