# Question cd3c6

Feb 3, 2018

It will be a $.1 \text{ M}$ solution of both $N {a}^{+}$ and $C {l}^{-}$ ions.

#### Explanation:

By salt, I assume you mean common table salt, $N a C l$. You are given that you have .2 moles of salt, thus you have .2" mol " Na^+ "and " .2 " mol " Cl^-.

Molarity is defined as the moles of a substance per Liter, so the molarity of the solution for both $N {a}^{+}$ and $C {l}^{-}$ is as follows:

(.2" moles")/(2" Liters") = .1 ("mol")/("Liter") = .1 " M"

Feb 3, 2018

${\text{0.1 mol L}}^{- 1}$

#### Explanation:

All you need to know here is that molarity is defined as the number of moles of solute present for every $\text{1 L}$ of the solution.

This means that in order to find the molarity of a given solution, you must determine how many moles of solute are present for every $\text{1 L}$ of this solution.

In your case, you know that $0.2$ moles of solute are dissolved in $\text{2 L}$ of the solution. Your goal here is to figure out how many moles of solute must be present in $\text{1 L}$ of this solution in order to have a solution of equal concentration to the one that contains $0.2$ moles of solute in $\text{2 L}$ of the solution.

$\text{? moles solute"/"1 L solution" = "0.2 moles solute"/"2 L solution}$

Rearrange to get

? = (1 color(red)(cancel(color(black)("L solution"))))/(2color(red)(cancel(color(black)("L solution")))) * "0.2 moles solute"

? = color(darkgreen)(ul(color(black)("0.1 moles solute")))

The answer is rounded to one significant figure.

So, you can say that this solution will have a molarity of ${\text{0.1 mol L}}^{- 1}$, which implies that $\text{1 L}$ of the solution will contain $0.1$ moles of solute.

Feb 3, 2018

The molarity of the salt solution is $\text{0.1 M}$.
"molarity"=("moles of solute")/("1 L of solution")#
There are $\text{0.2 mol of salt}$ dissolved in $\text{2 liter of salt solution}$.
$\text{molarity"=("0.2 mol salt")/("2 L")="0.1 mol salt/L = 0.1 M salt}$