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to prepare 300.0 mL of 0.2 M NaCl, how many grams of NaCl should be weighed out?
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use the molarity formula: \(M=\dfrac{n_{solute}}{L_{solution}}\) you can solve for moles then convert to mass, or substitute in \(n=\dfrac{m}{M_{molar}}\) \(M=\dfrac{\dfrac{m}{M_{molar}}}{L_{solution}}=\dfrac{m}{M_{molar}*L_{solution}}\) Rearranges to: \(m=M*M_{molar}*L_{solution}\) Where: \(M_{molar}\)=molar mass m=mass L=liters of solution n=moles M=molarity
@aaronq awesome work as usual!!!!!!!!
thank you!!!
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