Question #f5c04
1 Answer
Explanation:
You're interested in finding the volume of water needed to reduce the concentration of
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
Rearrange the above equation to solve for
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
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
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
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")