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

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

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Answer 1 · 2016-08-26T01:13:23

Easy.

Explanation:

$2.5$ $2.5$ $l i t e r s$ $liters$ of a $0.08$ $0.08$ $m o l a r$ $molar$ solution of $H C I$ $HCI$ gives us $m o l$ $mol$ $e s$ $es$ of $H C l$ $HCl$ in the solution

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

$0.08$ $0.08$ $(mol)/cancel L$ * $2.5 cancel L$ = $0.2 mol$ $es$ $HCl$

Now you must convert $mol$ $es$ to $grams$

$Molar$ $mass$ of $HCl$
$H = 1.01 (g)/(mol)$
$Cl = 35 (g)/(mol)$

$HCl = 1+35=36(g)/(mol)$

$(0.2 cancel mol)/1$ * $36 (g)/(cancel mol)$ = $color (red) (7.2 g)$ $HCl$

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

1 Answer

Explanation:

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