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

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

The component is present in greater amount

The component is present in lesser amount.

The mixture of solvent and solute

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

**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 can be expressed in different units.

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

- 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}

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

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 totoal 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

molar mass of NaCl = 23 *1 + 35.5*1

molar mass of NaCl = 58.5 g mol^{-1}

Amount = mass / molar mass

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

Dissolved NaCl amount = 0.1 mol

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.

Substitute NaCl moles and volume of the solution.

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

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 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.

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

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

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

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.

M_{Na2CO3} = 23*2 + 12 + 16*3

M_{Na2CO3} = 106 g mol^{-1}

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

Amount of Na_{2}CO_{3} = 0.1 mol

M_{K2CO3} = 39*2 + 12 + 16*3

M_{K2CO3} = 138 g mol^{-1}

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

Amount of K_{2}CO_{3} = 0.1 mol

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.

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}

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}

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 pOH from pH by using
the **pH + pOH = 14 **(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

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.

pH + pOH = 14

13 + pOH = 14

pOH = 1

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}

Concentration, moles amount, and volume relation is used.

C = n/v

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

n = 0.05mol

Multiply required moles times molecular weight

m = mass

n = moles

M = molecular weight

n = m/M

m = M*n

m = 0.05 * 40

m = 2g

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