# Question 6d652

Sep 30, 2017

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 $\text{1 L}$ of the solution.

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

${c}_{1} = \frac{n}{V} _ 1$

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 40%. If you take ${c}_{2}$ to be the new molarity of the solution, you can say that

${c}_{2} = \frac{140}{100} \cdot {c}_{1}$

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

${c}_{2} = \frac{n}{V} _ 2$

Here ${V}_{2}$ is the volume of the second solution.

This means that you have

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}}}{V} _ 2 = \frac{140}{100} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}}}{V} _ 1$

This is equivalent to

$\frac{1}{V} _ 2 = \frac{7}{5} \cdot \frac{1}{V} _ 1$

Rearrange to solve for ${V}_{2}$

${V}_{2} = \frac{5}{7} \cdot {V}_{1}$

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

${V}_{1} - \frac{5}{7} \cdot {V}_{1} = \frac{2}{7} \cdot {V}_{1}$

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.