Question

Question #0d05c

Answer 1

`"0.145 M"`

Explanation:

Before doing any calculation, try to predict what you expect the new concentration to be compared with the initial one.

As you know, molarity tells you how many moles of solute, which in your case is sodium chloride, `"NaCl"`, you get per liter of solution.

You can thus find the molarity of a solution by dividing the number of moles of solute by the volume of the solution expressed in liters

`color(blue)("molarity" = "moles of solute"/"liters of solution")`

Now, you can have two things happen when the number of moles of solute is kept constant.

• you can decrease the concentration of the solution by increasing its volume `->` make it more dilute

• you can increase the concentration of the solution by decreasing its volume `->` make it more concentrated

Notice that the volume of your solution decreases due to the evaporation of water. The number of moles of sodium chloride remain unchanged, but the volume of the solution decreases `->` you can expect the concentration to increase.

Think of it like this - you have the same amount of solute and less solvent.

So, use the molarity and volume of the initial solution to figure out how many moles of sodium chloride it contains

`color(blue)(c = n_"solute"/V_"solution" implies n_"solute" = c xx V_"solution")`

`n_(NaCl) = 0.130"moles"/color(red)(cancel(color(black)("L"))) * 375 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.04875 moles NaCl"`

The second solution will contain the same number of moles of sodium chloride, which means that you can use its volume to find its new molarity

`c_2 = "0.04875 moles"/(337 * 10^(-3)"L") = color(green)("0.145 M")`

The answer is rounded to three sig figs.

Indeed, the predict turned out to be correct, the concentration of the second solution is higher than that of the first solution.