What occurs to the concentration of a salt solution of $79 \cdot m L$ $79mL$ volume if $10 \cdot m L$ $10mL$ of solution are discarded?

Question

What occurs to the concentration of a salt solution of $79 \cdot m L$ $79mL$ volume if $10 \cdot m L$ $10mL$ of solution are discarded?

2 Answers

Impact of this question

1798 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-22T16:50:52

You only change the volume of the solution.....

Explanation:

By definition, $concentration \equiv \frac{moles of solute}{volume of solution}$ $"concentration"-="moles of solute"/"volume of solution"$ ....

And after the solution has been prepared it is homogeneous, and has the SAME CONCENTRATION whatever the volume.

And so remaining are $79 \cdot m L$ $79*mL$ of precisely the same concentration as the $10 \cdot m L$ $10*mL$ that were discarded. Of course you reduce the molar quantity, but NOT its concentration.

Answer 2 · 2017-10-22T16:52:28

See below.

Explanation:

Keep this formula in mind;

$M V = n$ $MV=n$ ,

where $M$ $M$ is the molarity of solution in $\frac{mol solute}{Liters solution} (\frac{m o l}{L})$ $"mol solute"/"Liters solution " ((mol)/L)$ , $V$ $V$ is the volume in $Liters (L)$ $"Liters " (L)$ , and $n$ $n$ is the amount of solute in $moles (m o l)$ $"moles "(mol)$ .

Also, know this formula:

$M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1=M_2V_2$ ,

where the terms on the left represent the initial values, and the terms on the right are the calculated final volume and molarity values.

--

Now, let's begin with the actual solution. Since we are given that $10$ $10$ mL are lost from the initial to final volume, let's plug the given values into the second equation mentioned above:

$M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1=M_2V_2$

$(0.20 M) (89 mL) = (x) (89 - 10 mL)$ $(0.20" M")(89" mL")=(x)(89-10" mL")$

Now, our job is to solve for $x$ $x$ , the new concentration.

$x = \frac{(0.20 M) (89 mL)}{89 - 10 mL}$ $x=((0.20" M")(89" mL"))/(89-10" mL")$

$x = 0.22 M$ $x=0.22" M"$

Therefore, the concentration of the solution $increases$ $color(red)("increases")$ .

Now, let's see about the moles:

$n = M V$ $n=MV$

First, we will be calculating the initial amount of moles $NaOH$ $"NaOH"$ present in solution:

$n = (0.20 M) (89 mL \cdot \frac{1 L}{10^{3} mL})$ $n=(0.20" M")(89" mL "*(1" L")/(10^3" mL"))$

$n = 0.0178 mol$ $color(blue)(n=0.0178" mol")$

Now, let's calculate the mole amount after the $10 mL$ $10" mL"$ has been removed:

$n = M V$ $n=MV$

$n = (0.22 M) (89 - 10 mL \cdot \frac{1 L}{10^{3} mL})$ $n=(0.22" M")(89-10" mL" * (1" L")/(10^3" mL"))$

$n = 0.174 mol$ $color(blue)(n=0.174" mol")$

Therefore, as you can see comparing the two blue values, the number of moles in solution actually $increased$ $color(red)("increased")$ slightly from the original solution after the $10 mL$ $10" mL"$ was removed.

I hope that helps!

Science

Math

Humanities

... and beyond

What occurs to the concentration of a salt solution of $79 \cdot m L$ $79mL$ volume if $10 \cdot m L$ $10mL$ of solution are discarded?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What occurs to the concentration of a salt solution of 79*mL79⋅mL79*mL volume if 10*mL10⋅mL10*mL of solution are discarded?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

What occurs to the concentration of a salt solution of $79 \cdot m L$ $79mL$ volume if $10 \cdot m L$ $10mL$ of solution are discarded?