Question

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

Answer 1

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

Explanation:

By definition, `"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*mL` of precisely the same concentration as the `10*mL` that were discarded. Of course you reduce the molar quantity, but NOT its concentration.

Answer 2

See below.

Explanation:

Keep this formula in mind;

`MV=n`,

where `M` is the molarity of solution in `"mol solute"/"Liters solution " ((mol)/L)`, `V` is the volume in `"Liters " (L)`, and `n` is the amount of solute in `"moles "(mol)`.

Also, know this formula:

`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` mL are lost from the initial to final volume, let's plug the given values into the second equation mentioned above:

`M_1V_1=M_2V_2`

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

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

`x=((0.20" M")(89" mL"))/(89-10" mL")`

`x=0.22" M"`

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

Now, let's see about the moles:

`n=MV`

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

`n=(0.20" M")(89" mL "*(1" L")/(10^3" mL"))`

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

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

`n=MV`

`n=(0.22" M")(89-10" mL" * (1" L")/(10^3" mL"))`

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

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

I hope that helps!