How do you calculate the mass of HCI required to prepare 2.5 liters of a 0.08 molar solution of HCI?

Aug 26, 2016

Easy.

Explanation:

$2.5$ $l i t e r s$ of a $0.08$ $m o l a r$ solution of $H C I$ gives us $m o l$$e s$ of $H C l$ in the solution

$M o l a r i t y$*$V o l u m e$ = $m o l$$e s$

$0.08$ $\frac{m o l}{\cancel{L}}$ * $2.5 \cancel{L}$ = $0.2 m o l$$e s$ $H C l$

Now you must convert $m o l$$e s$ to $g r a m s$

$M o l a r$ $m a s s$ of $H C l$
$H = 1.01 \frac{g}{m o l}$
$C l = 35 \frac{g}{m o l}$

$H C l = 1 + 35 = 36 \frac{g}{m o l}$

$\frac{0.2 \cancel{m} o l}{1}$ * $36 \frac{g}{\cancel{m} o l}$ = $\textcolor{red}{7.2 g}$ $H C l$