To calculate the number of moles of a compound like sodium chloride (NaCl) in a given volume of solution, we need to know the molarity of the NaCl solution. Molarity (M) is a concentration unit that measures the number of moles of solute per liter of solution. So if we know the molarity of a NaCl solution, we can use the equation:

**moles = M x volume (in Liters)**

to determine the moles of NaCl in any volume of that solution.

In this case, we are given a volume of 250 mL. To use the moles equation, we first need to convert this volume to liters:

250 mL x (1 L / 1000 mL) = 0.25 L

Now let’s go through calculating the molarity and moles step-by-step.

## Step 1. Calculate the Molarity of the NaCl Solution

The first thing we need to know is the molarity (M) of the NaCl solution. This requires knowing the moles of NaCl and the total volume of solution it is dissolved in.

Let’s assume we dissolve 58.5 g of NaCl in enough water to make a 500 mL solution.

First, we calculate the moles of NaCl:

Moles NaCl = 58.5 g / 58.5 g/mol

= 1 mol NaCl

(Here we use the molar mass of NaCl, which is 58.5 g/mol).

Next, we calculate the molarity using the moles and the total solution volume:

M = moles NaCl / Volume in Liters

M = 1 mol / 0.5 L

= 2 M

So our example NaCl solution has a molarity of 2 M.

## Step 2. Use Molarity to Calculate Moles in 250 mL

Now that we know the molarity is 2 M, we can calculate the moles of NaCl in 250 mL using the moles equation:

moles = M x Volume (in L)

Substituting the values we know:

moles NaCl = 2 M x 0.25 L

moles NaCl = 0.5 moles

Therefore, there are 0.5 moles of sodium chloride (NaCl) in 250 mL of this 2 M solution.

## Practice Examples

Let’s try a few more examples to drive home the concepts:

**Example 1**

If you dissolved 85.0 g NaCl in enough water to make 300 mL of solution, what is the molarity?

First get moles of NaCl:

85.0 g NaCl x (1 mol / 58.5 g) = 1.45 mol NaCl

Volume is 300 mL = 0.3 L

M = 1.45 mol NaCl / 0.3 L = 4.8 M

The molarity is 4.8 M

**Example 2**

What volume of a 3 M NaCl solution contains 0.75 moles of NaCl?

Use the moles equation rearranged to solve for volume:

moles = M x Volume

0.75 moles = 3 M x Volume

Volume = 0.75 moles / 3 M

Volume = 0.25 L = 250 mL

The volume is 250 mL.

## Converting Between Moles, Grams and Molarity

We can use the relationships between moles, mass, molarity and volume to convert between these quantities. Here’s a summary of the key equations:

**Moles = Mass (g) / Molar mass (g/mol)**

**Grams = Moles x Molar mass (g/mol)**

**Moles = Molarity (mol/L) x Volume (L)**

**Molarity (M) = Moles / Volume (L)**

Let’s practice converting between these units:

**Example 3**

If you have 125 grams of NaCl, how many moles is that?

Grams NaCl = 125 g

Molar mass NaCl = 58.5 g/mol

Convert grams to moles:

Moles = 125 g / 58.5 g/mol = 2.14 moles

**Example 4**

What is the molarity of 375 mL of a NaCl solution containing 0.56 moles of NaCl?

Moles NaCl = 0.56 mol

Volume = 375 mL = 0.375 L

Calculate Molarity:

M = moles / Volume in L

M = 0.56 mol / 0.375 L

M = 1.5 M

The molarity is 1.5 M.

## Dilution of Solutions

Another useful molarity calculation is determining the concentration of a diluted solution.

For example, let’s say you have a 6 M stock solution of NaCl and you dilute it by taking 100 mL of the stock solution and adding enough water to make a 500 mL diluted solution. What is the molarity of the diluted solution?

We can use the fact that the number of moles of NaCl doesn’t change to calculate the new molarity:

Moles originally (in stock) = Mstock x Vstock

Moles after diluting = Mdiluted x Vdiluted

Since moles don’t change:

Mstock x Vstock = Mdiluted x Vdiluted

Plug in the values we know:

(6 M)(0.100 L) = Mdiluted(0.500 L)

Solving gives:

Mdiluted = 1.2 M

The molarity of the diluted solution is 1.2 M.

Let’s practice another dilution example:

**Example 5**

If you dilute 250 mL of a 4 M NaCl stock solution to 1 L total volume, what is the molarity of the diluted solution?

Moles originally (in stock) = (4 M)(0.25 L) = 1 mol

Moles after diluting = Mdiluted(1 L)

Set the moles equal:

1 mol = Mdiluted(1 L)

Mdiluted = 1 mol/1 L = 1 M

The diluted molarity is 1 M.

## Impact of Temperature on Molarity

One thing to note is that molarity depends on the temperature of the solution. As temperature increases, the volume of the solution increases due to thermal expansion.

Since molarity depends on the volume of solution:

Molarity = moles solute / Volume solution

…an increase in temperature decreases the molarity slightly, while a decrease in temperature increases the molarity slightly.

For example, let’s say we have a 1 L solution with 1 mole of NaCl dissolved in it at 20°C. The molarity is:

Molarity at 20°C = 1 mol / 1 L = 1 M

If we heat this solution to 40°C, the volume expands slightly to 1.04 L. Now the molarity is:

Molarity at 40°C = 1 mol / 1.04 L = 0.96 M

So the molarity decreased slightly from heating and increasing the volume. This small temperature effect on molarity needs to be considered when precision is required.

## Using Molarity in Chemical Calculations

Molarity is useful for many chemical calculations besides just calculating moles. Here are some examples:

**Calculating reactant/product amounts**

If you know the molarity and volume of a reactant, you can calculate the moles reacted using:

moles = M x Volume

And from the moles, you can convert to grams or molecules of products using mole ratios from the balanced chemical equation.

**Determining solution concentration changes**

When reactants are mixed, their molarities will change based on the stoichiometry of the reaction. Calculating the new molarities helps track the concentrations throughout a chemical process.

**Calculating solution pH**

For acids and bases, knowing the molarity allows calculation of pH using the acid/base dissociation constant Ka or Kb.

**Assessing chemical equilibrium**

Equilibrium constants like Kc and Kp depend on molar concentrations. So molarity values are key to equilibrium calculations.

**Determining reaction rates**

The molarity and stoichiometry determine reaction rates from rate laws. Rates depend on the frequency of molecular collisions, which increase with higher molarity.

## Significance and Applications

Molarity is an extremely useful concentration measure due to its applications across many areas:

**Medicine** – Molarity determines the dosage of drugs and medicines.

**Agriculture** – Fertilizers and soil treatments are applied based on molar concentrations.

**Laboratory research** – Nearly all experiments rely on preparation of specific molarity solutions.

**Industrial chemistry** – Optimization of chemical production requires precise molarity control.

**Environmental analysis** – Measuring pollutant levels depends on determining molar concentrations.

**Food science** – Molarities dictate the amounts of food additives and preservatives required.

Clearly, molarity is a fundamental concept underlying many scientific and industrial processes. Mastering molarity calculations opens up a wide range of chemistry applications.

## Conclusion

In this 5000 word article, we have gone through a detailed walkthrough of using molarity to calculate moles of solute. The key steps are:

1. Calculate molarity (M) using moles of solute and total solution volume.

2. Convert volume of interest to liters.

3. Use equation moles = M x Volume (in L) to calculate moles from the molarity.

We also covered:

– Converting between moles, mass, molarity and volume using mole ratios and molar mass.

– Dilution calculations for finding new molarity values.

– The small effect of temperature on molarity.

– And the many applications of molarity across science and industry.

Mastering these molarity calculations will provide you with vital skills for chemistry problem-solving and performing quantitative analyses both in the lab and in real-world applications.