Question #f5c04

1 Answer
Dec 29, 2015

Answer:

#"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")#