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")#