Question

How many milliliters of
11.5 M HCl(aq)
are needed to prepare
490.0 mL
of
1.00 M HCl(aq)

Answer 1

$M1*V1=M2*V2$

$V1=M2*V2/V1=490mL*1M/11.5M=42.61mL$

Answer:42.61mL