Question

How many mL of solvent are required to make a 3.0 M NaCl solution for a reaction that requires 146.3 g of NaCl?

Answer 1

ANSWER

The volume of the solution is 833.3mL

EXPLANATION

Given that;

The molarity of NaCl is 3.0M

The mass of NaCl is 146.3 grams

Follow the steps below to find the volume of the solution

Step 1; Find the number of moles of solute using the below formula

$\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of NaCl is 58.44 g/mol

$\begin{gathered} \text{ mole = }\frac{\text{ 146.3}}{\text{ 58.44}} \\ \text{ mole= 2.50 moles} \end{gathered}$

Step 2; Find the volume of the solution using the molarity

$\text{ Molarity = }\frac{\text{ no of moles }}{\text{ volume of the solution}}$$\begin{gathered} \text{ 3 = }\frac{\text{ 2.50}}{\text{ V}} \\ \text{ cross multiply} \\ \text{ V}\times\text{ 3 = 2.50} \\ \text{ 3V = 2.50} \\ \text{ Divide both sides by 3} \\ \text{ }\frac{\text{ 3V}}{3}\text{ }=\text{ }\frac{\text{ 2.5}}{3} \\ \text{ V = 0.833L} \end{gathered}$

Step 3; Convert the volume from L to mL

1mL is equivalent to 0.001L

Let x represents the volume of the solution mL

$\begin{gathered} \text{ 1mL }\rightarrow\text{ 0.001L} \\ \text{ xmL }\rightarrow\text{ 0.833L} \\ \text{ cross multiply} \\ \text{ 1mL }\times\text{ 0.833L }=\text{ xmL }\times\text{ 0.001L} \\ \text{ Isolate x} \\ \text{ x }=\text{ }\frac{\text{ 1mL}\times0.833\cancel{L}}{0.001\cancel{L}} \\ \text{ x = }\frac{0.833}{0.001} \\ \text{ x = 833.3mL} \end{gathered}$

Therefore, the volume of the solution is 833.3mL