# Concentration Calculation Questions, Answers | Molarity Examples

In this tutorial, we study about molarity and molality theories of solutions and solve questions to get
better understand.

**Content**

In this tutorial, we study about followings of sodium metal

**Definition of concentration**
- Solvent, solute, solution definition

**Molarity of a solution**
- Molarity equation
- Molarity units
- How to calculate concentration in mol dm
^{-3}?
- Examples about molarity values
- Calculating molarity of solutions

## Definition of concentration

Concentration tells us, the amount of solute present in a given quantity of solvent or solution is expressed in
terms of concentration. We can express concentration in different ways as below. units of those parameters
are noted.

- Mass percentage (m/m%) or volume percentage (v/v%) - No units
- Molarity of solution - mol dm
^{-3}
- Molality of solution - mol kg
^{-1}
- Mole fraction - No units

### Solvent, solute, solution definition

Understanding the definitions and meaning of is compulsory to solve problems of concentrations

**What is solvent?**

The component is present in greater amount

**What is the solute?**

The component is present in lesser amount.

**What is solution?**

The mixture of solvent and solute

In this tutorial we try to study the **molarity** and **molality** of solutions.

## Molarity of a solution

**The number of moles of the solute dissolved per unit volume of the solution** is the definition of
molarity. Units can be mol dm^{-3} or mol m^{-3}.

### Molarity equation

### Molarity units

Molarity can be expressed in different units.

- mol dm
^{-3} : mostly used
- mol l
^{-}
- kmol m
^{-3}

### How to calculate concentration in mol dm^{-3} ?

In the units, there are mol and dm^{-3}. They tells us about amount and volume respectively. Therefore we have
to find the dissolved amount and volume of the solution to calculate concentration in mol dm^{-3}.

### Examples about molarity values

- If 1 dm
^{3} aqueous solution contains 0.45 mol of HCl, molarity of HCl is 0.45 mol dm^{-3}.
Otherwise, we can describe this is as, 0.45 mol of HCl is dissolved 1 dm^{3} of the total solution.
- If 10 dm
^{3} aqueous solution contains 0.98 mol of HNO_{3}, molarity of HNO_{3} is 0.098 mol dm^{-3}.

Sometimes we use '[ solute ]' to represent concentration of solute. As an example, to represent
concentration of aqueous HCl, we can use [HCl_{(aq)}]

Example: [HNO_{3(aq)}] = 0.1 mol dm^{-3}

### Calculating molarity of solutions - Example calculations

Before calculating molarity of solutions, you should have an idea about following parametres.

- Calculating
**molar mass** when relative atomic masses are known
- Relationship of molar mass (M), mass (m) and amount (n): n = m/M

#### Calculate molarity - Example 1

5.85 g of NaCl is dissolved in 500cm^{3} of distilled water. Calculate
the molarity of NaCl mol dm^{-3}.
( Na = 23 , Cl = 38.5 )

Answer

Dissolved amount (moles) and total volume of the solution are required to calculate the concentration according to the
molarity equation. In this example, total solution volume is given as a data. Therefore we have to find the dissolved
NaCl amount.

Mass of dissolved NaCl is given. Now we can **find the NaCl amount** using **mass, molar mass and amount relationship**.
First **molar mass of NaCl** should be calculated. Let's calculate the molar mass of NaCl.

**Find the molar mass of NaCl**

- molar mass of NaCl = 23 *1 + 35.5*1
- molar mass of NaCl = 58.5 g mol
^{-1}

**Find the amount of moles of dissolved NaCl**

Amount = mass / molar mass

- Dissolved NaCl amount = 5.85 g / 58.5 g mol
^{-1}
- Dissolved NaCl amount = 0.1 mol

**Calculate the volume of total solution in dm**^{3}

- 1dm
^{3} = 1000cm^{3}
- 500cm
^{3} = 500 * 10^{-3 }dm^{3} = 0.5dm^{3}

Now, we know the volume of solution in dm^{3} and amount of moles of NaCl. So we can find molarity of NaCl now.

#### Molarity Equation

Substitute NaCl moles and volume of the solution.

Molarity of NaCl = 0.2 mol dm^{-3}

#### Calculate molarity - Example 2

You are advised to prepare 0.47 mol dm^{-3} NaOH solution in the laboratory.
Solid NaOH, distilled water and other lab facilities are supplied for you. If you have to
prepare 25cm^{3} solution, calculate followings.

- Required amount of moles of NaOH
- Required NaOH mass

In this example, required molarity 0.47 mol dm^{-3} and volume ( 25cm^{3}) are given. When we know the molarity and volume,
we can easily calculate the required amount of moles of NaOH using relationship of concentration, moles, and volume
of solution. After calculating NaOH moles, required NaOH mass is calculated.

**Calculate moles of NaOH**

- Concentration = Moles / Volume
- Moles of NaOH = Concentration * volume
- Moles of NaOH = 0.47 mol dm
^{-3} * 25 * 10^{-3} dm^{3}
- Moles of NaOH = 0.01175 mol

**Calculate required mass of NaOH**

Now we can use the **relation of mass, molar mass and amount of moles** to find required mass of NaOH.

- Molar Mass of NaOH = 23 + 16 + 1
- Molar Mass of NaOH = 40 g mol
^{-1}

- Amount = Mass / Molar Mass
- Required NaOH mass = NaOH amount * Molar Mass of NaOH
- Required NaOH mass = 0.01175 mol * 25 * 40 g mol
^{-1}
- Required NaOH mass = 0.47 g

#### Calculate molarity - Example 4

A mixture of Na_{2}CO_{3} and K_{2}CO_{3}
is dissolved in distilled water upto 250 cm^{3}. Dissolved mixture includes
10.6 g of Na_{2}CO_{3} and 13.8 g of K_{2}CO_{3}.

- Calculate amount of moles each Na
_{2}CO_{3} and K_{2}CO_{3} compounds.
- Express concentrations in mol dm
^{-3} of each Na_{2}CO_{3} and K_{2}CO_{3}.
- Calculate concentrations of CO
_{3}^{2-} in the solution.

K=39 , Na=23 , O=16 , C=12

Step 1: Masses of Na_{2}CO_{3} and K_{2}CO_{3} are given as data. So we can find amount of
Na_{2}CO_{3} and K_{2}CO_{3}. In this example, We use 'M' to express molar mass as
a symbol.

**Find molar masses of Na**_{2}CO_{3}

- M
_{Na2CO3} = 23*2 + 12 + 16*3
- M
_{Na2CO3} = 106 g mol^{-1}

#### Find amount of Na_{2}CO_{3}

- Amount of Na
_{2}CO_{3} = 10.6 g / 106 g mol^{-1}
- Amount of Na
_{2}CO_{3} = 0.1 mol

#### Find molar masses of K_{2}CO_{3}

- M
_{K2CO3} = 39*2 + 12 + 16*3
- M
_{K2CO3} = 138 g mol^{-1}

#### Find amount of K_{2}CO_{3}

- Amount of K
_{2}CO_{3} = 13.8 g / 138 g mol^{-1}
- Amount of K
_{2}CO_{3} = 0.1 mol

Step 2: Now we know the amount of moles of each Na_{2}CO_{3} and K_{2}CO_{3}.
Therefore molarity can be found now.

#### Find molarity

**Molarity = Dissolved Amount / Volume of Solution**

- [Na
_{2}CO_{3(aq)}] = 0.1 mol / 0.25 dm^{3}
- [Na
_{2}CO_{3(aq)}] = 0.4 mol dm^{-3}

- [K
_{2}CO_{3(aq)}] = 0.1 mol / 0.25 dm^{3}
- [K
_{2}CO_{3(aq)}] = 0.4 mol dm^{-3}

Step 3: CO_{3}^{2-} ions are given from both Na_{2}CO_{3} and K_{2}CO_{3} compounds.
Both Na_{2}CO_{3} and K_{2}CO_{3} dissociate completely to ions in the water. Therefore
total amount of CO_{3}^{2-} ions in the water equals to the sum of CO_{3}^{2-} given by
Na_{2}CO_{3} and CO_{3}^{2-} given by K_{2}CO_{3}. After calculating
total CO_{3}^{2-}, molarity is calculated.

- Total CO
_{3}^{2-} amount = CO_{3}^{2-} from Na_{2}CO_{3} +
CO_{3}^{2-} from K_{2}CO_{3}
- Total CO
_{3}^{2-} amount = 0.1 + 0.1
- Total CO
_{3}^{2-} amount = 0.2 mol

- Therefore [CO
_{3}^{2-}] = 0.2 mol / 0.25 dm^{3}
- Therefore [CO
_{3}^{2-}] = 0.8 mol dm^{-3}

#### Calculate molarity - Example 5

You have to prepare a NaOH solution and its pH value should be 13. Total volume of
the solution is 500 ml. Calculate the required mass of NaOH to prepare this solution at 25^{0}C.

**Calculate concentration of a pH known solution**

- Calculate pOH from pH by using the
**pH + pOH = 14 ** equation (at 25^{0}C). Then you can calculate the concentration of
NaOH solution by **pOH = -log**_{10}[OH^{-}_{(aq)}] .
- Then find the required moles of NaOH by the equation of
**C=n/v** . Here C= concentration, n=required moles,
V - volume of solution
- Now weight is measured by multiplying n and molar mass.

**Assumptions**

The OH^{-} received from water dissociation is negligible when it compare with OH^{-} received
from NaOH.

**Calculate pOH from pH value**

- pH + pOH = 14
- 13 + pOH = 14
- pOH = 1

**Calculate concentration(mol dm**^{-3})of solution

Now we know the pOH of NaOH solution. Use the relation of pOH and OH^{-} concentration

- pOH = -log
_{10}[OH^{-}_{(aq)}]
- 1 = -log
_{10}[OH^{-}_{(aq)}]
- [OH
^{-}_{(aq)}] = 0.1 mol dm^{-3}

**Find the required moles(mol) of NaOH**

Concentration, moles amount, and volume relation is used.

C = n/v

- 0.1 = n/0.5 [500ml = 0.5 dm
^{3}]
- n = 0.05mol

### Find the required weight(g)

Multiply required moles by molecular weight to find the required weight

- m = mass
- n = moles
- M = molecular weight
- n = m/M

- mass = Molecular weight * moles
- m = 0.05 * 40
- m = 2g

### Molarity of distilled water

Now we are going to discuss about how to calculate molarity of water in mol dm^{-3}. Take the density of water as 1000 g dm^{-3}.

First calculate the mass of water in 1 dm^{3}

mass = volume * density

mass = 1 dm^{3} * 1000 g dm^{-3}

mass = 1000 g

Next, find how many moles are in 1dm^{3} of distilled water.

Number of moles = mass / molecular weight

Number of moles = 1000 g / 18 g mol^{-1}

Moles = 55.56 mol

Finally, calculate the concentration (molarity)

Concentration = dissolved moles / total volume

Concentration = 55.56 mol / 1 dm^{3}

Concentration of water = 55.56 mol dm^{-3}

### How to calculate concentration in mol dm^{-3}?

From mol dm^{-3} we know, we have to calculate the molarity. This means how many moles of solvent are dissolved in
total of 1 dm^{3} volume of **solution**.

Example 5

### Calculate concentration of aqueous Na_{2}CO_{3} in mol dm^{-3} .

In a Na_{2}CO_{3} solution, concentration of Na^{+} is found that 0.008 mol dm^{-3}. Calculate the concentration of Na_{2}CO_{3} in mol dm^{-3}.

Na_{2}CO_{3} hydrolysis is occurred as followings in the water.

Na_{2}CO_{3} → 2Na^{+} + CO_{3}^{2-}

According to the stoichiometry ratio, one Na_{2}CO_{3} mol gives two Na^{+} moles.

So concentration of Na_{2}CO_{3} is half of the concentration of Na^{+} ions.

Concentration of Na_{2}CO_{3} = 0.008 / 2

Concentration of Na_{2}CO_{3} = 0.004 mol dm^{-3}

Your Questions

### 25cm^{3} of X_{2}CO_{3} contains 0.106 g required 23 cm^{3} and 0.217 molarity of HCl acid
for complete neutralization. Calculate molar mass of X_{2}CO_{3}

We need to neutralize CO_{3}^{2-}. So first, we need to write the chemically balanced equation to solve this problem.

X_{2}CO_{3} + 2HCl = 2XCl_{2} + CO_{2} + H_{2}O

#### Stoichiometry of the equation

1 mol X_{2}CO_{3} = 2 mol HCl

- Amount of spent HCl = 0.217 mol dm
^{-3} * 23 * 10^{-3} dm^{-3}
- Amount of spent HCl = 0.004991 mol

According to the Stoichiometry required amount of HCl is twice as amount of X_{2}CO_{3}. Therefore existed amount of X_{2}CO_{3} is,

- Existed X
_{2}CO_{3} amount = 0.004991 mol / 2
- Existed X
_{2}CO_{3} amount = 0.0024955 mol

### We dissolve 2 moles of salt in 1dm^{3} of water to make a solution. Find the molar concentration

This is a straight forward question. You are provided two required data to find the concentration.

- Amount of moles = 2 moles
- Volume of the solution = 1 dm
^{3}

Then substitute the values to concentration equation.

- Concentration = amount / volume
- Concentration of salt = 2 mol / 1 dm
^{3}
- Concentration of salt = 2 mol dm
^{-3}

### What is the molarity of 25 cm^{3} of a solution of 5.85 g?

How do you calculate molarity if you do not know the compound because number of moles are required to calculate molarity. So molar mass
should be known. Molar mass differs from compound to compound.

### molarity equation a level chemistry?

Molarity = dissolved amount (mol) / volume of solution (dm^{3})

### How to calculate molarity questions given the amount of moles and volume

This is a very easy question. Divide number of moles by volume of solution.

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