Mole and solutions
REMEMBER: WHEN working problems with moles, it is always best to start by finding the number of moles of the known substance, whether by using its mass or volume. Also, in most cases, an equation is essential as it helps to determine in what mole ratio the reactants combine or the products form.
A solution is made up of a solute dissolved in a solvent.
The concentration of a solution is the amount of solute dissolved in a given volume of solvent.
Concentration can be expressed as molar concentration or mass concentration.
Molar concentration is the moles of solute in a given volume of solvent (mol dm-3).
Mass concentration is the mass of solute in a given volume of solvent (g dm-3).
A standard solution is a solution for which the concentration is known.
The concentration (C) can be used to determine the number of moles (n) in the volume (V) of that solution (n= C (mol dm-3) * V (dm3))
Remember 1dm3 = 1litre = 1,000cm3; Molar (M) represents mol dm-3
EXAMPLE 1
How many moles of HCl are present in 250cm3 of a 2mol/dm3 solution? Concentration =2 M = 2mol in 1,000cm3 HCl Therefore, X mol would be in 250cm3 Cross multiply, X = (2*250)/ 1,000 = 0.5mol HCl
EXAMPLE 2
What is the concentration in mol dm-3 of 49 g of H2SO4 dissolved in 250cm3 of solution?
Determine the number of moles in 49g of H2SO4
Mass of 1 mole H2SO4 = ((1*2) + 32 + (16*4)) = 98g
Number of moles in 49g H2SO4 = 49/98 = 0.5mol
So 0.5mol H2SO4 is in 250 cm3
Therefore, X mol are in 1,000cm3 (this is the concentration) X = (0.5*1,000)/250 = 2mol Concentration is 2mol/ dm3 H2SO4
EXAMPLE 3
Find the mass concentration (g/ dm3) if 0.6 mol HNO3 is dissolved in 2 dm3 of solution. Method 1: 0.6 mol HNO3 in 2 dm3 Thus X mol in 1 dm3 X = (0.6*1)/2 = 0.3mol HNO3 Molar mass HNO3 =(1 + 14 + (16*3)) = 63g mol-1 Mass of 0.3mol = 0.3*63 = 18.9g HNO3 Mass concentration = 18.9g/ dm3 HNO3
METHOD 2:
Molar mass of HNO3 = (1+14 + (16*3)) = 63g/mol Mass of 0.6 mol = 0.6*63 = 37.8g HNO3 So 37.8g HNO3 in 2dm3 Xg in 1 dm3; X = (37.8*1)/ 2 = 18.9g HNO3 Mass concentration = 18.9g/ dm3 HNO3
EXAMPLE 4
Calculate the mass of NaOH required to prepare 500 cm3 of a 0.1mol dm-3 solution.
Determine the number of moles to be placed in 500cm3
Concentration = 0.1 M = 0.1mol NaOH in 1000cm3 Therefore we have X mol NaOH in 500cm3 Cross multiply to solve for X X = (0.1*500)/ 1,000 = 0.05mol NaOH Molar mass of NaOH = 23 + 16 + 1 = 40g mol-1
Mass of 0.05mol NaOH = 0.05*40 = 2g NaOH
EXAMPLE 5
Calculate the mass of sodium carbonate needed to make up 500cm3 of a 0.20M solution.
M (molarity) is the number of moles in 1dm3, that is mol/dm3. Thus, 0.2M = 0.2mol in 1,000cm3 (1dm3) X mol == 500cm3 X = (500x0.2)/1,000 = 0.1 mol Mr of Na2CO3 = 106g, thus the mass of 0.1 mol = 106x0.1 = 10.6g 10.6g of Na2CO3 dissolved in 500cm3 has a concentration of 0.2M
ALTERNATIVE METHOD
What mass of NaOH is needed to make up 250cm3 of a 2M solution? 2M = 2mol in 1,000cm3 (1dm3) Mr of NaOH = 40g then 2mol == 80g 80g = 1,000cm3 Xg = 250cm3 X = (250x80)/1,000 = 20g of NaOH
CONCENTRATION CALCULATION
As stated earlier, concentration can be expressed in mol/dm3 and g/dm3.
EXAMPLE 6
What is the concentration of 20cm3 of sulphuric acid containing 0.25 mol H2SO4? 0.25mol == 250 cm3 x mol == 1,000cm3 (1dm3) x = (0.25x1,000)/250 = 1mol Concentration = 1 mol/dm3
Calculate the concentration in g/dm3 of the same solution. Since, 0.25 mol H2SO4 is present in 250cm3 Mr of H2SO4 = 98g mass of 0.25mol = 98x0.25 = 24.5g Thus, 24.5g === 250cm 3 Xg ==== 1,000cm3 X = (1,000x24.5)/250 = 98g: hence, Concentration = 98g/dm3 Alternative method Since the concentration = 1 mol/dm3 Concentration in g/dm3 = 1 mol/dm3 x Mr = 1 mol/dm3 x 98 g/mol = 98g/dm3
Further Practice: Find the concentration in g/dm3 and mol/dm3 of the following solutions
(i) 30 cm3 of a nitric acid solution containing 0.10mol HNO3
(ii) 200cm3 of a sodium hydroxide solution containing 40g of NaOH
(iii) 40cm3 of an ammonium nitrate solution containing 16g of NH4NO3
(iv) 350cm3 of a sulphuric acid solution containing 0.185mol H2SO4. THE REAL TEST OF THIS TOPIC COMES IN ITS APPLICATION TO TITRATION PROBLEMS!!