How do you calculate the concentration of solute?
Answer
608.1k+ 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.
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