
How do you calculate the concentration of solute?
Answer
547.8k+ views
Hint: Concentration is defined as ratio of amount of solute to amount of solution [It is a homogeneous mixture containing solute and solvent, where the solvent is major component of solution and solute is minor component and active ingredient.] Solute is minor forms like gas, liquid or solid.
Complete step by step answer:
There are many ways of expressing concentration:
Parts per million (PPm)
Percentage by weight (%W/W)
Percentage by volume. (%V/V)
Molality $(\text{m})$
Molarity $(\text{M})$
The most commonly used and accurate method to find concentration is ‘’Molarity’’ and ‘’Molality’’
Molality $(\text{m})$: It is a measure of number of moles of solute present in 1 kg of solvent.
$\text{m}=\dfrac{\text{moles of solvent}}{\text{mass of solution}}=\dfrac{\text{n (moles)}}{\text{kg}}$
Molarity/Mole concentration: It is defined as the number of moles dissolved in 1L solution.
$\text{M=}\dfrac{\text{moles solutes}}{\text{volume of solution}}=\dfrac{\text{n (moles)}}{\text{v}\,\text{(L)}}$
Unit $=\text{M,}\ \text{mol/L}$
Mole: It is a unit which is used to quantitatively measures the amount of substance
$1\ \text{mole}=6.022\times {{10}^{23}}$ particles or molecules of substance
$6.022\times {{10}^{23}}=$ Avogadro number
Mole of any substance is calculated by dividing the mass of that substance by the molar mass (MM) of that molecule or atom in gm.
Example: A $20.0\ \text{mL}$ of water contain 10.5 gram of glucose $(\text{MM}=\text{180}\text{.18}\ \text{g/mol}).$ The molarity of solution will be:
Convert g of glucose to mol of glucose
$10.5\ \text{g glucose}\times \dfrac{\text{1 mol glucose}}{\text{180}\text{.18 g glucose}}=0.0583\ \text{mol}$
Convert ml to L
$\text{20}\text{.0 mL}\times \dfrac{\text{0}\text{.001}\ \text{L}}{1\,\text{mL}}=0.0200\ \text{L}$
Molarity $=\dfrac{\text{0}\text{.0583 mol glucose}}{0.0200\ \text{L}}=2.92\ \text{M}$
Note: Colligative properties are properties of solutions that are affected by the number of particles. E.g: Boiling point elevation, Freezing point depression and osmotic pressure.
Colligative properties depend on concentration of solute but not depend upon the identity of solute.
Complete step by step answer:
There are many ways of expressing concentration:
Parts per million (PPm)
Percentage by weight (%W/W)
Percentage by volume. (%V/V)
Molality $(\text{m})$
Molarity $(\text{M})$
The most commonly used and accurate method to find concentration is ‘’Molarity’’ and ‘’Molality’’
Molality $(\text{m})$: It is a measure of number of moles of solute present in 1 kg of solvent.
$\text{m}=\dfrac{\text{moles of solvent}}{\text{mass of solution}}=\dfrac{\text{n (moles)}}{\text{kg}}$
Molarity/Mole concentration: It is defined as the number of moles dissolved in 1L solution.
$\text{M=}\dfrac{\text{moles solutes}}{\text{volume of solution}}=\dfrac{\text{n (moles)}}{\text{v}\,\text{(L)}}$
Unit $=\text{M,}\ \text{mol/L}$
Mole: It is a unit which is used to quantitatively measures the amount of substance
$1\ \text{mole}=6.022\times {{10}^{23}}$ particles or molecules of substance
$6.022\times {{10}^{23}}=$ Avogadro number
Mole of any substance is calculated by dividing the mass of that substance by the molar mass (MM) of that molecule or atom in gm.
Example: A $20.0\ \text{mL}$ of water contain 10.5 gram of glucose $(\text{MM}=\text{180}\text{.18}\ \text{g/mol}).$ The molarity of solution will be:
Convert g of glucose to mol of glucose
$10.5\ \text{g glucose}\times \dfrac{\text{1 mol glucose}}{\text{180}\text{.18 g glucose}}=0.0583\ \text{mol}$
Convert ml to L
$\text{20}\text{.0 mL}\times \dfrac{\text{0}\text{.001}\ \text{L}}{1\,\text{mL}}=0.0200\ \text{L}$
Molarity $=\dfrac{\text{0}\text{.0583 mol glucose}}{0.0200\ \text{L}}=2.92\ \text{M}$
Note: Colligative properties are properties of solutions that are affected by the number of particles. E.g: Boiling point elevation, Freezing point depression and osmotic pressure.
Colligative properties depend on concentration of solute but not depend upon the identity of solute.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

