Question

Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20 g/mL.


How much concentrated solution would you take to prepare 3.05L of 0.480M HCl by mixing with water? In ML

Answer 1

Note % as mass
\[Number of Moles=Mass(g)/Molar Mass(g/mol) \]\[Number of moles of HCl=37/36.5=1.01 \]Density=Mass(g)/Volume(ml) 100/V=1.20 1.20V=100 V=83.3 ml change to L by dividing by 1000
\[Molarity = moles / Volume = 1.01 mol / 0.0833 L = 12.12 M \]so you take a solution of a known concentration you pour more of HCL what happens? Dilution, you changed the concentration by changing the volume what you haven't changed is the number of moles of solute that's in there. the changed is that you added more volume.
Use Dilution Law
\[C(intial)*Volume(Intial)=C(final)*V(final) \]\[12.12*V(intial)=3.05*0.480 \]\[V(intial)=3.05*0.480/12.12=0.120 L \]
Change 0.120 L to ml by *1000=120 ml

Answer 2

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