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How many milliliters of 11.5 M HCl(aq) are needed to prepare 935.0 mL of 1.00 M HCl(aq)?
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You need to use the dilution equation, where you have two sets of conditions: \(\sf \huge \underbrace{M_1V_1}_{have }=\underbrace{M_2V_2}_{sought}\) \(\sf 11.5 M *V_1=935.0 mL* 1.00 M \rightarrow V_1=\dfrac{935.0 mL* 1.00 M}{11.5 M }\)
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