Final answer:
The mass of bromine in 1 dm³ of Dead Sea water, based on a bromide ion concentration of 0.0526 mol/dm³, is calculated to be 8.40668 g/dm³. None of the answer choices a) 159.80 g b) 79.90 g c) 39.95 g d) 19.98 g match this calculated value.
Step-by-step explanation:
The question is asking for the mass of bromine in 1 dm³ of Dead Sea water based on the bromide ion concentration.
To find the mass, we use the molar mass of bromine (Br️) which is approximately 79.90 g/mol. Since the concentration is given as 0.0526 mol/dm³ for bromide ions and each bromide ion weighs about 79.90 g/mol, the mass of bromine in 1 dm³ of Dead Sea water can be calculated by multiplying the concentration by the molar mass of bromine (mass = concentration × molar mass).
To calculate: 0.0526 mol/dm³ × 79.90 g/mol = 4.20334 g/dm³. However, because each molecule of bromine (Br️) consists of two bromine atoms, we must consider this in calculating the mass of bromine molecules. Therefore, the mass of bromine will be twice the mass of a single bromine atom: 4.20334 g/dm³ × 2 = 8.40668 g/dm³.
The closest answer from the options provided is 8.39 g, which is not explicitly listed among the choices (a) 159.80 g (b) 79.90 g (c) 39.95 g (d) 19.98 g. None of the given options are correct based on our calculation. Therefore, a clarification or re-evaluation of the question and its choices may be necessary.