# Question #a14ed

##### 1 Answer

#### Explanation:

The most important thing to remember when dealing with a **dilution** is that the *number of moles of solute*, i.e. the amount of solute, you have in the initial solution **remains constant** after the dilution is performed.

In essence, diluting a solution implies **increasing** its volume **while** keeping the number of moles of solute constant.

Since molarity is defined as the number of moles of solute *per liter of solution*, increasing the volume while keeping the number of moles of solute constant will result in a **decrease** in concentration.

#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liters of solution"color(white)(a/)|)))#

In this case, you know that the *diluted solution* has a molarity of **every** liter of solution contains

So, you know that this solution contains

Since you performed a *dilution*, you can say for a fact that this is how many moles of sulfuric **must have been present** in the initial solution.

This means that the initial solution had a molarity of

#c_"initial" = "2.68 moles"/(318 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"8.43 mol L"^(-1)color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.

This is exactly what you get by using the equation for **dilution calculations**, which looks like this

#color(blue)(overbrace(c_1 xx V_1)^(color(black)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(black)("moles of solute in diluted solution"))#

Here you have

If you rearrange the equation to solve for

#c_1V_1 = c_2V_2 implies c_1 = V_2/V_1 * c_2#

#c_1 = (1.00 color(red)(cancel(color(black)("L"))))/(318 * 10^(-3)color(red)(cancel(color(black)("L")))) * "2.68 M" = color(green)(|bar(ul(color(white)(a/a)"8.43 mol L"^(-1)color(white)(a/a)|)))#

It's worth noting that the ratio between the **final volume** and the **initial volume** of the solution is called the **dilution factor**.

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"final"/V_"initial"color(white)(a/a)|)))#