Iron II Bromide"= FeBr_3" Molar mass of= "295.56 g\/mol"
"moles = 16.3 g \\times \\frac{1 mol}{295.56 g} = 0.056moles"
liters of the solution = 375 ml x 1 L/1000 ml = 0.375 L
"Molarity=\\frac{Moles}{Volume}=\\frac{0.056}{0.375}=0.147M"
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