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# How many grams of ammonium sulfate are needed to make a $0.25{\text{ }}L$ solution at a concentration of $6{\text{ }}M$ ?

Last updated date: 22nd Feb 2024
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Hint: We realize that the concentration alludes to the measure of solute which disintegrates in the unit measure of solution/solvent. Here concentration of solution can be spoken to in different structures, for example, measure of solute disintegrates in given mass/volume of solvent just as measure of solute present in given sum volume/mass of a solution.

Concentration of solution is likewise characterized as the amount of solute which is available in a specific amount of a solution/solvent. In this manner, there are different various techniques to characterize concentration of an answer like Molarity, Molality. Likewise, the most well-known unit among concentration terms is molarity. It's much more significant just as valuable for figuring stoichiometry of response in solution. It's characterized as number of moles of a solute which is available in $1{\text{ }}litre$ of solution as it's given as:
Your procedure here will be to utilize the volume and molarity of the answer for decide the number of moles of ammonium sulfate, ${\left( {N{H_4}} \right)_2}S{O_4}$ , it should contain.
When you know the quantity of moles of solute expected to make that solution, use ammonium sulfate's molar mass to decide the number of grams would contain that numerous moles.
The equation is given by;
$concentration\; = \;\dfrac{{mole\;of\;solute}}{{volume\;of\;solution}}$
Since, $mass\;of\;solute\; = \;moles\; \times \;molar\;mass$
Consequently, subbing the qualities we get;
$= {\text{ }}Volume{\text{ }} \times {\text{ }}Concentration{\text{ }} \times {\text{ }}Molar{\text{ }}Mass{\text{ }}of{\text{ }}Ammonium{\text{ }}Sulfate$
$= 0.25\not{{{\text{L solution}}}} \cdot \dfrac{{{\text{6 }}\not{{{\text{moles}}{{\left( {{\text{N}}{{\text{H}}_4}} \right)}_2}{\text{S}}{{\text{O}}_4}}}}}{{1\not{{{\text{L solution}}}}}}.\dfrac{{{\text{132}}{\text{.14 g}}}}{{1\not{{{\text{mole}}{{\left( {{\text{N}}{{\text{H}}_4}} \right)}_2}{\text{S}}{{\text{O}}_4}}}}}$ (where, $132.14{\text{ }}g{\text{ }}mo{l^{ - 1}}$ is the molar mass of ammonium)
$\Rightarrow {\text{198}}{\text{.21 g}}$
In this manner, ${\text{198}}{\text{.21}}$ grams of ${\left( {{\text{N}}{{\text{H}}_4}} \right)_2}{\text{S}}{{\text{O}}_4}$ are expected to plan $0.25L$ $6M$ Ammonium sulfate.
Adjusted to one critical figure, the appropriate response will be
${\text{mass of ammonium sulfate}} = \left[ {200g} \right]$
Accordingly, the concentration can be report on mass to mass $\left( {\dfrac{m}{m}} \right)$ premise or as we can say that on mass to volume $\left( {\dfrac{m}{v}} \right)$ premise just as this is mostly use in designing applications and in a clinical lab.

Note: Note that the concentration of solution is one of a type of naturally visible property. It additionally can be communicated in different structures like both amount savvy just as quality astute. Along these lines’ quality can be quality can communicate as weak solution or centralization of solution just as the amount astutely it very well may be communicated as immersed or as unsaturated.