Difference between revisions of "Molarity"
(→Equation) |
|||
| Line 27: | Line 27: | ||
: volume = The [[Volume (Space)|volume]] of [[solvent]]. | : volume = The [[Volume (Space)|volume]] of [[solvent]]. | ||
| − | + | ===Calculating Molarity=== | |
{| class="wikitable" | {| class="wikitable" | ||
|- | |- | ||
| style="height:20px; width:300px; text-align:center;" |117g of [[NaCl]] is [[dissolve]]d in 0.5dm<sup>3</sup> of [[water]]. Calculate the [[concentration]] in g/dm<sup>3</sup>. | | style="height:20px; width:300px; text-align:center;" |117g of [[NaCl]] is [[dissolve]]d in 0.5dm<sup>3</sup> of [[water]]. Calculate the [[concentration]] in g/dm<sup>3</sup>. | ||
| style="height:20px; width:300px; text-align:center;" |28g of [[KOH]] is [[dissolve]]d in 100ml of [[water]]. Calculate the [[concentration]] in g/dm<sup>3</sup>. | | style="height:20px; width:300px; text-align:center;" |28g of [[KOH]] is [[dissolve]]d in 100ml of [[water]]. Calculate the [[concentration]] in g/dm<sup>3</sup>. | ||
| − | | style="height:20px; width:300px; text-align:center;" | | + | | style="height:20px; width:300px; text-align:center;" |7.3g of [[HCl]] is [[dissolve]]d in 400ml of [[water]]. Calculate the [[concentration]] in g/dm<sup>3</sup>. |
|- | |- | ||
| style="height:20px; width:200px; text-align:left;" | | | style="height:20px; width:200px; text-align:left;" | | ||
| Line 65: | Line 65: | ||
| style="height:20px; width:200px; text-align:left;" | | | style="height:20px; width:200px; text-align:left;" | | ||
'''Find the number of [[mole]]s.''' | '''Find the number of [[mole]]s.''' | ||
| − | [[mass]] = | + | [[mass]] = 7.3g |
[[Relative Formula Mass]] = 36.5g | [[Relative Formula Mass]] = 36.5g | ||
| Line 73: | Line 73: | ||
Number of [[Mole]]s = <math>{\frac{m}{M_r}}</math> | Number of [[Mole]]s = <math>{\frac{m}{M_r}}</math> | ||
| − | Number of [[Mole]]s = <math>{\frac{ | + | Number of [[Mole]]s = <math>{\frac{7.3}{36.5}}</math> |
| − | Number of [[Mole]]s = 2mol | + | Number of [[Mole]]s = 0.2mol |
|- | |- | ||
| style="height:20px; width:300px; text-align:left;" | | | style="height:20px; width:300px; text-align:left;" | | ||
| Line 118: | Line 118: | ||
'' '''concentration (mol/dm<sup>3</sup>)''' '' = <math>\frac{mass (g)}{volume (dm^3)}</math> | '' '''concentration (mol/dm<sup>3</sup>)''' '' = <math>\frac{mass (g)}{volume (dm^3)}</math> | ||
| − | '' '''concentration''' '' = <math>\frac{2}{0.4}</math> | + | '' '''concentration''' '' = <math>\frac{0.2}{0.4}</math> |
| − | '' '''concentration''' '' = 5mol/dm<sup>3</sup> = 5M | + | '' '''concentration''' '' = 0.5mol/dm<sup>3</sup> = 5M |
|} | |} | ||
Latest revision as of 11:34, 12 January 2019
Contents
Key Stage 3
Meaning
Molarity is aunit of concentration of solutions.
About Molarity
- Molarity is shortened with an upper case M.
- When molarity is used to describe a solution it may be referred to as molar, eg. a 0.5 molar solution of Sodium Chloride.
- Bottles containing acid and alkali usually have their Molarity stated on the side. The more concentrated the higher the molarity.
- The molarity of Stomach Acid is around 0.16 Molar (0.16M).
- The molarity of Battery Acid can be as much as 5.2 Molar (5.2M).
Key Stage 4
Meaning
Molarity is a unit of concentration of solutions in mol/dm3.
About Molarity
- Molarity is shortened with an upper case M.
- Molarity is the ratio of the number of moles of a solute to the volume of solvent.
- When molarity is used to describe a solution it may be referred to as molar, eg. a 0.5 molar solution of Sodium Chloride.
- Bottles containing acid and alkali usually have their Molarity stated on the side. The more concentrated the higher the molarity.
Equation
concentration (mol/dm3) = \(\frac{Moles (mol)}{volume (dm^3)}\)
Where:
- Moles = The number of moles of the solute. Found by dividing the mass in grams by the relative formula mass.
- volume = The volume of solvent.
Calculating Molarity
| 117g of NaCl is dissolved in 0.5dm3 of water. Calculate the concentration in g/dm3. | 28g of KOH is dissolved in 100ml of water. Calculate the concentration in g/dm3. | 7.3g of HCl is dissolved in 400ml of water. Calculate the concentration in g/dm3. |
|
Find the number of moles. mass = 117g Relative Formula Mass = 58.5g Number of Moles of a Compound = (Mass of compound)/(Relative Formula Mass of compound) Number of Moles = \({\frac{m}{M_r}}\) Number of Moles = \({\frac{117}{58.5}}\) Number of Moles = 2mol
|
State the mass in grams and the volume in dm3: Find the number of moles. mass = 28g Relative Formula Mass = 56g Number of Moles of a Compound = (Mass of compound)/(Relative Formula Mass of compound) Number of Moles = \({\frac{m}{M_r}}\) Number of Moles = \({\frac{28}{56}}\) Number of Moles = 0.5mol |
Find the number of moles. mass = 7.3g Relative Formula Mass = 36.5g Number of Moles of a Compound = (Mass of compound)/(Relative Formula Mass of compound) Number of Moles = \({\frac{m}{M_r}}\) Number of Moles = \({\frac{7.3}{36.5}}\) Number of Moles = 0.2mol |
|
volume = 0.5dm3 |
volume = 100ml The volume must be converted into dm3 from ml. volume in dm3 = (volume in ml)/1000 volume = 0.1dm3 |
volume = 400ml The volume must be converted into dm3 from ml. volume in dm3 = (volume in ml)/1000 volume = 0.4dm3 |
|
Calculate the Concentration: concentration (mol/dm3) = \(\frac{Moles (mol)}{volume (dm^3)}\) concentration = \(\frac{2}{0.5}\) concentration = 4mol/dm3 = 4M |
Calculate the Concentration: concentration (mol/dm3) = \(\frac{Moles (mol)}{volume (dm^3)}\) concentration = \(\frac{0.5}{0.1}\) concentration = 5mol/dm3 = 5M |
Calculate the Concentration: concentration (mol/dm3) = \(\frac{mass (g)}{volume (dm^3)}\) concentration = \(\frac{0.2}{0.4}\) concentration = 0.5mol/dm3 = 5M |