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.
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.
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.
Sometimes we use '[ solute ]' to represent concentration of solute. As an example, to represent concentration of aqueous HCl, we can use [HCl(aq)]
Example: [HNO3(aq)] = 0.1 mol dm-3
Before calculating molarity of solutions, you should have an idea about following parametres.
5.85 g of NaCl is dissolved in 500cm3 of distilled water. Calculate the molarity of NaCl mol dm-3. ( Na = 23 , Cl = 38.5 )
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
1dm3 = 1000cm3
500cm3 = 500 * 10-3 dm3 = 0.5dm3
Now, we know the volume of solution in dm3 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 25cm3 solution, calculate followings.
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 dm3
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 Na2CO3 and K2CO3 is dissolved in distilled water upto 250 cm3. Dissolved mixture includes 10.6 g of Na2CO3 and 13.8 g of K2CO3.
K=39 , Na=23 , O=16 , C=12
Masses of Na2CO3 and K2CO3 are given as data. So we can find amount of Na2CO3 and K2CO3. In this example, We use 'M' to express molar mass as a symbol.
MNa2CO3 = 23*2 + 12 + 16*3
MNa2CO3 = 106 g mol-1
Amount of Na2CO3 = 10.6 g / 106 g mol-1
Amount of Na2CO3 = 0.1 mol
MK2CO3 = 39*2 + 12 + 16*3
MK2CO3 = 138 g mol-1
Amount of K2CO3 = 13.8 g / 138 g mol-1
Amount of K2CO3 = 0.1 mol
Now we know the amount of moles of each Na2CO3 and K2CO3. Therefore molarity can be found now.
Molarity = Dissolved Amount / Volume of Solution
[Na2CO3(aq)] = 0.1 mol / 0.25 dm3
[Na2CO3(aq)] = 0.4 mol dm-3
[K2CO3(aq)] = 0.1 mol / 0.25 dm3
[K2CO3(aq)] = 0.4 mol dm-3
CO32- ions are given from both Na2CO3 and K2CO3 compounds. Both Na2CO3 and K2CO3 dissociate completely to ions in the water. Therefore total amount of CO32- ions in the water equals to the sum of CO32- given by Na2CO3 and CO32- given by K2CO3. After calculating total CO32-, molarity is calculated.
Total CO32- amount = CO32- from Na2CO3 + CO32- from K2CO3
Total CO32- amount = 0.1 + 0.1
Total CO32- amount = 0.2 mol
Therefore [CO32-] = 0.2 mol / 0.25 dm3
Therefore [CO32-] = 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 250C.
Calculate pOH from pH by using
the pH + pOH = 14 (at 250C)
Then you can calculate the concentration of NaOH solution by pOH = -log10[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.
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 = -log10[OH-(aq)]
1 = -log10[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 dm3]
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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