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How many moles of ions are present in aqueous solutions prepared by dissolving 10.00g of the f scandium III bromide in water to make a 1.25 L solution?
Wiki User
∙ 14y ago
First figure out the formula of Scandium(III) Bromide: its ScBr3. Next ignore the 1.25 L solution part -- it's just extra information. Use the 10.00g to convert it to moles. To do so first figure out the molar mass of ScBr3: its 285g/mol. Now you do: 10.00g x 1 mol/285g = .0350877193... mol. Next round it to 3 significant figures: .0351. Now you find out the number of ions in the formula and use it as a conversion factor (instead of Avagrado's number, 6.02 x 1023): it has 4 ions (Scandium = 1; Bromide = 3). Finally you do: .0351 mol x 4 mol ions/1 mol = .1404 mol ions. This is your answer! By Geovonni Bell ;-)
Wiki User
∙ 14y ago
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