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# Molarity Formula: Sample Problems with Solutions and a Quiz (Complete Tutorial)

## First, what is molarity?

Molarity is one of the several ways to describe the concentration of a solution. It is basically how many moles of the solute are present in your solution.

You can calculate the Molarity of your solution using this formula:

**M = n/V**

Where M is the Molarity, n is the number of moles of the solute and V is the volume of your solution.

The chemicals in your laboratory have specific Molarity. Hence, your teacher will insist that you learn to compute the Molarity of your solution. He or she might even ask to prepare a solution with a certain Molarity.

## Molarity Formula in Action: Sample Problem # 1

*What is the molarity of a 2 L solution containing 20 g of sodium chloride?*

To solve this problem, we must list down the given quantities first based on the Molarity formula.

M = n/V

Given:

Volume of solution (V) = 2 L

Moles of Solute (n) =* not given*

Molarity =* not **given*

There are two unknowns in the Molarity formula: moles of solute and the Molarity. Since we are asked for the Molarity, we should look for the moles of solute first.

The problem tells us that there is 25 g of NaCl in the solution. Using your previous knowledge about molar mass, we can calculate the number of moles of the solute using dimensional analysis.

Let us calculate the molar mass of NaCl first.

Na › 23 g

Cl › 35 g

Molar Mass = 23 g + 35 g = 58 g

Then, let’s solve for the moles of NaCl.

25 g NaCl x 1 mole NaCl / 58 g NaCl =0.43 moles NaCl

We now have the number of moles of the solute. Finally, we can solve for the Molarity of this solution.

M = n/V

M = 0.43 moles NaCl / 2 L

M = 0.22 moles NaCl/L

M = 0.22 M

*Therefore, the Molarity of the solution is 0.22.*

## Molarity Formula in Action: Sample Problem # 2

Let’s try another Molarity problem but with a shorter solution.

*What is the Molarity of a half liter solution containing 5 g of copper sulfate?*

M = n/V

Given:

Moles of solute = *unknown*

Volume of solution = 0.5 L

Mass of solute (CuSO4) = 5 g

Cu › 64 g

S › 32 g

O › 16 g x 4 = 64 g

Molar Mass of CuSO_{4} = 64 g + 32 g + 64 g = 160 g

5 g CuSO_{4} x 1 mole CuSO_{4} / 160 g CuSO_{4 }= 0.03 moles CuSO_{4}

M = n/V

M = 0.03 moles CuSO_{4} / 0.5 L

M = 0.6 M

*Therefore, the Molarity of the solution is 0.06.*

## How to Determine the Amount of Solute Using the Molarity Formula

Okay, number three in the above quiz was a little bit tricky. I haven’t taught you that one yet.

If you got it right, you are very good! If not, don’t worry because I will teach you how to solve it.

The third problem is just the opposite of the first two sample problems. Let’s take a look at the problem again.

*How many grams of NaCl is present in 2 liters of a 0.05 M brine solution?*

First, identify what you can arrive with the given using the Molarity formula.

M = n/V

Given:

M = 0.05 M

V = 2 L

n =* unknown*

Let’s now solve for the number of moles of solute (n).

0.05 M = n/2 L

N = 0.1 moles NaCl

Now that we have the number of moles of NaCl, let’s use it to calculate the mass of NaCl in the solution.

0.1 moles NaCl x 58 g NaCl / 1 mole NaCl = 5.8 g NaCl

*Therefore, there are 5.8 grams of NaCl in the solution.*

**Questions?** Write a comment below and I'll answer as soon as I can.

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