Question #a667b
1 Answer
Explanation:
Your strategy here will be to use the molarity and volume of the initial solution to determine how many moles of solute, which in your case is sodium hydroxide,
The idea is that the number of moles of solute will remain constant, since you're only evaporating water, your solvent.
The concentration of the solution will increase because after some of the water is evaporates, the same number of moles of solute will not occupy a smaller volume.
Molarity is defined as moles of solute per liter of solution, which means that you have
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#
The initial solution will thus contain
#n_("NaOH") = "6.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * 2.348color(red)(cancel(color(black)("L"))) = "14.088 moles NaOH"#
Convert the volume of evaporated water from milliliters to liters by using
#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(a/a)|)))#
You will have
#761 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.761 L"#
The volume of water that remains after evaporating the known
#V_"remaining" = "2.348 L" - "0.761 L" = "1.587 L"#
The molarity of the new solution will be
#c_"new" = "14.088 moles"/"1.587 L" = color(green)(|bar(ul(color(white)(a/a)"8.9 M"color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the concentration of the initial solution.
As predicted, the molarity of the solution increased as a result of the decrease in the volume of solvent.
