Question

Question #f5c04

Answer 1

`"1070 mL"`

Explanation:

You're interested in finding the volume of water needed to reduce the concentration of `"800.0 mL"` of water from `"1.75 M"` to `"0.750 M"`.

As you know, molarity is defined as moles of solute divided by liters of solution.

`color(blue)("molarity" = "moles of solute"/"liters of solution")`

The idea here is that adding water to an aqueous solution will keep the number of moles of solute constant, but will increase the volume of the solution - this is known as diluting a solution.

The molarity of the solution will decrease because the same number of moles of solute will now be distributed in a larger volume.

So, how many moles of solute are present in your initial solution?

`color(blue)(c = n/V implies n = c * V)`

`n = "1.75 M" * 800.0 * 10^(-3)"L" = "1.40 moles"`

Now the question becomes what volume of solution is needed in order for `1.40` moles of solute to correspond to a molarity of `"0.750 M"`.

Rearrange the above equation to solve for `V`, the volume of the target solution

`color(blue)(c = n/V implies V = n/c)`

`V = (1.40 color(red)(cancel(color(black)("moles"))))/(0.750color(red)(cancel(color(black)("moles")))/"L") = "1.8667 L"`

So, the target solution must have a volume of `"1.8667 L"`. This means that you must add

`V = V_"initial" + V_"water"`

`V_"water" = "1.8667 L" - 800.0 * 10^(-3)"L" = "1.0667 L"`

of water to the initial solution to get its concentration to drop from `"1.75 M"` to `"0.750 M"`.

Rounded to three sig figs, and expressed in milliliters, the answer will be

`V_"water" = color(green)("1070 mL")`

Alternatively, you can use the formula for dilution calculations

`color(blue)(c_1 * V_1 = c_2 * V_2)" "`, where

`c_1`, `V_1` - the molarity and volume of the initial solution
`c_2`, `V_2` - the molarity and volume of the target solution

In your case, you'd have

`V_2 = c_1/c_2 * V_1`

`V_2 = (1.75 color(red)(cancel(color(black)("M"))))/(0.750 color(red)(cancel(color(black)("M")))) * "800.0 mL" = "1866.7 mL"`

Once again, the volume of water needed will be

`V_"water" = "1866.7 mL" - "800.0 mL" = color(green)("1070 mL")`