How many liters of each solution do you need to get 3.0 mol of HCl?

a. 12.0 M HCl b. 2 M HCl c. 0.5 M HCl d. 0.010 M HCl

Mar 21, 2018

By definition...$\text{concentration"="moles of solute"/"volume of solution}$

Explanation:

And so to gets the $\text{volume}$..we arrange the given quotient..

$\text{volume of solution"="moles of solute"/"concentration}$

$a .$ And so if $3.0 \cdot m o l$ are required from $12.0 \cdot m o l \cdot {L}^{-} 1$ $H C l$...we take the quotient...

$\frac{3.0 \cdot m o l}{12.0 \cdot m o l \cdot {L}^{-} 1} = 0.250 \cdot L$ (mind you I don't think you can get $H C l$ at this concentration....)

$b .$ And if $3.0 \cdot m o l$ are required from $2.0 \cdot m o l \cdot {L}^{-} 1$ $H C l$...we take the quotient...

$\frac{3.0 \cdot m o l}{2.0 \cdot m o l \cdot {L}^{-} 1} = 1.50 \cdot L$

$c .$ And if $3.0 \cdot m o l$ are required from $0.5 \cdot m o l \cdot {L}^{-} 1$ $H C l$...we take the quotient...

$\frac{3.0 \cdot m o l}{0.50 \cdot m o l \cdot {L}^{-} 1} = 6.0 \cdot L$

$d .$ And if $3.0 \cdot m o l$ are required from $0.010 \cdot m o l \cdot {L}^{-} 1$ $H C l$...we take the quotient...

$\frac{3.0 \cdot m o l}{0.010 \cdot m o l \cdot {L}^{-} 1} = 300.0 \cdot L$

And clearly this last is an impratical proposition....