Question #f5c04

1 Answer
Dec 29, 2015

"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")