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Question

Question: How do you calculate the concentration of solute?...

How do you calculate the concentration of solute?

Explanation

Solution

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 (m)(\text{m})
Molarity (M)(\text{M})
The most commonly used and accurate method to find concentration is ‘’Molarity’’ and ‘’Molality’’
Molality (m)(\text{m}): It is a measure of number of moles of solute present in 1 kg of solvent.
m=moles of solventmass of solution=n (moles)kg\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.
M=moles solutesvolume of solution=n (moles)v(L)\text{M=}\dfrac{\text{moles solutes}}{\text{volume of solution}}=\dfrac{\text{n (moles)}}{\text{v}\,\text{(L)}}
Unit =M, mol/L=\text{M,}\ \text{mol/L}
Mole: It is a unit which is used to quantitatively measures the amount of substance
1 mole=6.022×10231\ \text{mole}=6.022\times {{10}^{23}} particles or molecules of substance
6.022×1023=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 mL20.0\ \text{mL} of water contain 10.5 gram of glucose (MM=180.18 g/mol).(\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 g glucose×1 mol glucose180.18 g glucose=0.0583 mol10.5\ \text{g glucose}\times \dfrac{\text{1 mol glucose}}{\text{180}\text{.18 g glucose}}=0.0583\ \text{mol}
Convert ml to L
20.0 mL×0.001 L1mL=0.0200 L\text{20}\text{.0 mL}\times \dfrac{\text{0}\text{.001}\ \text{L}}{1\,\text{mL}}=0.0200\ \text{L}
Molarity =0.0583 mol glucose0.0200 L=2.92 M=\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.