# Question #0d05c

##### 1 Answer

#### 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, **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 candecreasethe concentration of the solution byincreasingits volume#-># make it moredilute

you canincreasethe concentration of the solution bydecreasingits volume#-># make it moreconcentrated

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 **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.