# Question 4e511

Jun 12, 2015

You need 8 g of sodium chloride to make that solution.

#### Explanation:

A solution's molarity is defined as the number of moles of solute, in your case sodium chloride, divided by the volume of the solution - expressed in liters.

Since you know what the volume and the molarity of the target solution have to be, you can determine how many moles of sodium chloride you'd need by

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N a C l} = 0.2 \text{moles"/cancel("L") * 700 * 10^(-3)cancel("L") = "0.14 moles}$

Notice that I converted the volume from mL to L. In addition to that, I used the fact that ${\text{1 mL" = "1 cm}}^{3}$.

Now use sodium chloride's molar mass, which tells you what the mass of 1 mole of sodium chloride is, to determine how many grams would contain that many moles.

0.14cancel("moles") * "58.44 g"/(1cancel("mole")) = "8.18 g"#

Rounded to one sig fig, the number of sig figs you gave for the volume and molarity of the solution, the answer will be

${m}_{N a C l} = \textcolor{g r e e n}{\text{8g}}$

So, in order to make this solution, you add 8 g of $N a C l$ to enough water to make the final volume equal to 700 mL.