Question

Question #216ab

Answer 1

`"0.083 L"` of concentrated `12.0M` `HNO_3`

Explanation:

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

`color(white)(aaaaaaaaaaaaaaaa)` `M_1V_1 = M_2V_2`

Where
`M_1 = "molarity of solution 1"`
`V_1 = "volume of solution 1"`
`M_2 = "molarity 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"` exactly is,

`"Molarity" = "moles"/(1" Liter solution")`

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

`color(white)(aaaaaaaaaaaaaaaa)`

Using our equation from above, we can find the number of moles of nitric acid, `HNO_3`, from what we are given.

We will let `M_1` and `V_1` be `8.0 M` and `125" mL"` respectively. Make sure to convert `"mL"` to `"L"` first. For `M_2`, use the `12.0M`. Isolate `V_2` and solve.

• `M_1V_1 = M_2V_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