Answer :
Final answer:
The mass of barium iodide in 188 ml of a 0.532M solution is calculated to be approximately 39.11 grams using concepts of molarity, moles, and molar mass.
Explanation:
The question is asking to find out the mass of barium iodide in a solution, given a certain volume and a concentration. The question is a standard calculation involving molarity and volume. Molarity (M) is defined as the number of moles of solute per liter of solution. Therefore, you can calculate the number of moles of solute first using the formula M = n/V, where n is the number of moles and V is the volume in liters. We can convert 188 mL to liters (0.188 L) and we know the molarity is 0.532M, so the number of moles in the solution is 0.532M * 0.188 L = 0.10 moles of iodide ions. To find the mass of barium iodide, we multiply this number by the molar mass of barium iodide, which is roughly 391.136 g/mol (from barium's molar mass of 137.327 g/mol and iodine's molar mass of 253.809 g/mol). Thus, the mass of barium iodide is 0.10 moles * 391.136 g/mol = 39.11 g.
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