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If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required ?

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M_(A)V_(A)=M_(B)V_(B)\\(80.0)(4.00)=V_(B)(12.4)\\V_(B)=((80.0)(4.00))/(12.4) \approx \boxed{25.8 \text{ mL}}

If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required-example-1
User Shubham Chahal
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