# Question #6d652

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

As you know, the **molarity** of a solution tells you the number of moles of solute present **for every**

This implies that you can **increase** the molarity of a solution by **decreasing** the amount of solvent it contains.

Now, let's say that the molarity of the initial solution is

#c_1 = n /V_1#

Here

#n# is the number of moles of solute present in the solution#V_1# is the volume of the initial solution

After you remove some of the solvent, the molarity of the solution **increases** by

#c_2 = 140/100 * c_1#

Since the number of moles of solute remained **constant**, you can say that

#c_2 = n/V_2#

Here

This means that you have

#color(red)(cancel(color(black)(n)))/V_2 = 140/100 * color(red)(cancel(color(black)(n)))/V_1#

This is equivalent to

#1/V_2 = 7/5 * 1/V_1#

Rearrange to solve for

#V_2 = 5/7 * V_1#

You can thus say that the volume of the solution **decreased** by

#V_1 - 5/7 * V_1 = 2/7 * V_1#

which is equivalent to a **percent decrease** of

#"% decrease" = (2/7 * color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * 100% = color(darkgreen)(ul(color(black)(29%)))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you have one significant figure for the increase in molarity.

Notice that this is equivalent to saying that

#"% change" = (5/7 * color(red)(cancel(color(black)(V_1))) - color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * 100% = - 29%#

Here the *minus sign* symbolizes **percent decrease**.