# Question b889f

May 26, 2017

62.3 L of the new $H C l$ solution can be made.

#### Explanation:

For these type of calculations, we use:
$\frac{m o l}{v o l u m e} = M$

Now we have $170$ mL $11$M $H C l$ in stock. We first calculate how many mol $H C l$ that is. We rewrite the above formula:

$m o l = v o l u m e \times M$

Filling in the values and units gives us:
$m o l = 0.170 \left(\textcolor{red}{\cancel{\textcolor{b l u e}{L}}}\right) \times 11 \left(\textcolor{b l u e}{\frac{m o l}{\textcolor{red}{\cancel{\textcolor{b l u e}{L}}}}}\right)$
$m o l = 1.87 \textcolor{b l u e}{\textcolor{w h i t e}{a} m o l}$

This is the amount of mol we can use to create the $0.03$ M solution. We use the same formula again to calculate which volume we can create with this new concentration.

$v o l u m e = \frac{m o l}{M}$

Filling in the values and units gives us:

volume=(1.87 (color(red)(cancel(color(blue)(mol)))))/(0.03(color(red)(cancel(color(blue)(mol)))xxcolor(blue)(L^-1))#

$v o l u m e = 62.3 \textcolor{b l u e}{\textcolor{w h i t e}{a} L}$

Therefore we can make 62.3 L of the new 0.03 $H C l$ M solution.