How many milliliters of 11.5 M HCl(aq) are needed to prepare 485.0 mL of 1.00 M HCl(aq)?
回答 (1)
2016-10-24 8:40 am
C₁V₁ = C₂V₂
Volume of 11.5 M HCl needed, V₁ = C₂V₂/C₁ = (1.00 M) × (475.0 mL) / (11.5 M) = 41.3 mL
Volume of 11.5 M HCl needed, V₁ = C₂V₂/C₁ = (1.00 M) × (475.0 mL) / (11.5 M) = 41.3 mL
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