Question

Question #9cb3f

Answer 1

See details below.

Explanation:

Whenever you are given a volume with a given concentration, when you take the product of these two quantities, you will get moles of that particular entity.

In each of the different scenarios, you are given moles of the solute. Then you are asked to find the new molarity (concentration) when `250 "mL"` of `H_2O` is added.

`color(white)(aaaaaaaa)` Working example of 125 mL of 0.251 M HCl

First step: Convert `mL->Liters` (used for all volumes)

Conversion factor used

• `(given cancel "mL of solute")/(1)* (1*10^-3 L)/(cancel "mL") = "volume in Liters"`

Second step : Find moles

`(0.125 "Liter")/1*(0.251 "moles")/(1 "Liter") = 0.031 "moles HCl"`

When you add water, you are essentially diluting the solution. If we look at what molarity is,

`Molarity = "moles"/"Liter"`

you can clearly see the relationship between volume and molarity. When you `color(blue)("increase the volume")`, you `color(blue)("decrease the molarity")`, an inverse relationship. Since we are adding water, which will increase the volume, our molarity for the solution should go down.

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)`

To find the new molarity or concentration, first, add the volume of water, `0.250 "L"` with the volume of whatever you are working with. In our case, it's `HCl`, so `0.125 "L"`.

`color(white)(aaaaaaaaaaa)`New Volume : `0.250 "L" + 0.125 "L" = 0.375 "L"`

Take the number of moles of HCl and divide by new volume to get new molarity.

`color(white)(aaaaaaaaaaaaaaa)`New Molarity:

`(0.031 "moles HCl")/(0.375 "L") = 0.0826 "M HCl solution"`

Repeat this for every given problem.