Question
Question: How many grams of \[MgS{{O}_{4}}\] needs to prepare \[125mL\] \[0.200M\] magnesium sulfate?...
How many grams of MgSO4 needs to prepare 125mL 0.200M magnesium sulfate?
Solution
We know that the concentration refers to the amount of solute which dissolves in the unit amount of solution/solvent. Here concentration of solution can be represented in various forms such as amount of solute dissolve in given mass/volume of solvent as well as amount of solute present in given amount volume/mass of a solution.
Complete step-by-step answer: Concentration of solution is also defined as quantity of solute which is present in a particular quantity of a solution/solvent. Thus there are various different methods to define concentration of a solution like Molarity, Molality. Also the most common unit among concentration terms is molarity. It's much important as well as useful for calculating stoichiometry of reaction in solution. It’s defined as number of mole of a solute which is present in 1 litre of solution as it's given as:
The formula is given by; concentration = volume of solutionmole of solute
Since, mass of solute = moles × molar mass
Therefore, substituting the values we get;
= Volume × Concentration × Molar Mass of Magnesium Sulfate
=(125×10−3L)×(0.200Lmol)×(120.37molg)
⇒ 3.00925 grams
Therefore, 3.00925 grams of MgSO4 are needed to prepare 125mL\text{ }$$$$0.200M magnesium sulfate.
Thus, the concentration can be report on mass to mass (mm) basis or as we can say that on mass to volume (vm) basis as well as this is mainly use in engineering applications and in a clinical labs.
Note: Note that the concentration of solution is one of a type of macroscopic property. It also can be expressed in various forms like both quantity wise as well as quality wise. Thus quality can be quality can express as dilute solution or concentration of solution as well as the quantity wise it could be expressed as saturated or as unsaturated.