Question #216ab

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Question #216ab

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Answer 1 · 2017-04-14T01:05:40

$0.083 L$ $"0.083 L"$ of concentrated $12.0 M$ $12.0M$ $H N O_{3}$ $HNO_3$

Explanation:

This is a dilution chemistry problem. For problems like this one, use the following equation

$a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaa)$ $M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1 = M_2V_2$

Where
$M_{1} = molarity of solution 1$ $M_1 = "molarity of solution 1"$
$V_{1} = volume of solution 1$ $V_1 = "volume of solution 1"$
$M_{2} = molarity of solution 2$ $M_2 = "molarity of solution 2"$
$V_{2} = volume of solution 2$ $V_2 = "volume of solution 2"$

Whenever you dilute a concentrated stock solution, or for that matter, any solution, you are essentially changing the volume of that solution by adding water. If we look at what $Molarity$ $"Molarity"$ exactly is,

$Molarity = \frac{moles}{1 Liter solution}$ $"Molarity" = "moles"/(1" Liter solution")$

you can see when you increase the $volume$ $"volume"$ , you decrease the $molarity$ $"molarity"$ (concentration). In dilution cases, you are adding water so moles of the solute do not change, only the volume of the solution changes.

$a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaa)$

Using our equation from above, we can find the number of moles of nitric acid, $H N O_{3}$ $HNO_3$ , from what we are given.

We will let $M_{1}$ $M_1$ and $V_{1}$ $V_1$ be $8.0 M$ $8.0 M$ and $125 mL$ $125" mL"$ respectively. Make sure to convert $mL$ $"mL"$ to $L$ $"L"$ first. For $M_{2}$ $M_2$ , use the $12.0 M$ $12.0M$ . Isolate $V_{2}$ $V_2$ and solve.

$M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1 = M_2V_2$
$(0.125 L) (8.0 M) = (12.0 M) (V_{2})$ $(0.125L)(8.0M) = (12.0M)(V_2)$
$((0.125L)(8.0cancelM))/(12.0cancelM) = V_2$
$"0.083 L"= V_2$

Answer: 0.083 L

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Question #216ab

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