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Question: How many grams of \(CuS{{O}_{4}}\) would be dissolved in \(5L\) of \(0.5M\) \(CuS{{O}_{4}}\) solutio...

How many grams of CuSO4CuS{{O}_{4}} would be dissolved in 5L5L of 0.5M0.5M CuSO4CuS{{O}_{4}} solution?

Explanation

Solution

This question is based on the concept of molarity, if we are thorough with the term Molarity, we can easily attempt the question. Molarity is defined as the number of moles of solute dissolved per litre of the solution.

Complete step-by-step answer:
It is given in the question, that the molarity of the acid= 0.5M0.5M
And volume of solution= 5L5L
Also, from the definition on molarity we know that
Molarity(M)=Numberofmolesofsolutevolumeofsolution(inL)Molarity\,\left( M \right)=\dfrac{Number\,of\,moles\,of\,solute}{volume\,of\,solution\,\left( in\,L \right)}
From the above formulae, we can calculate the number of moles as shown below
Numberofmoles=Molarity(M)×VolumeofsolutionNumber\,of\,moles\,=\,Molarity\left( M \right)\,\times \,Volume\,of\,solution\,
Therefore, we can say that
Number of moles of CuSO4CuS{{O}_{4}} = 0.5×5=2.50.5\times 5=2.5moles
Now we know that numberofmoles=givenmassofcompoundmolarmassnumber\,of\,moles\,=\,\dfrac{given\,mass\,of\,compound}{molar\,mass}
In this question we need to find out the mass of the Copper sulphate. We can find the mass of the acid by modifying the above formula as
(massofCuSO4)=numberofmolesofCuSO4×molarmassofCuSO4(mass\,of\,CuS{{O}_{4}})=\,number\,of\,moles\,of\,CuS{{O}_{4}}\,\times \,molar\,mass\,of\,CuS{{O}_{4}}
Now, firstly let us find out the molar mass of CuSO4CuS{{O}_{4}}
We know that molar mass of Copper= 63.5g/mol63.5 g/mol
Molar mass of Sulphur = 32g/mol32g/mol
Molar mass of oxygen=16g/mol16 g/mol
Therefore, Molar mass of CuSO4CuS{{O}_{4}}= 63.5+32+64=159.6or16063.5+32+64=159.6\,or\,160g/mol
massofCuSO4=2.5×160=400g\therefore mass\,of\,CuS{{O}_{4}}=\,\,2.5\,\times \,160=\,400g
Hence, we can say that 160160 gram of CuSO4CuS{{O}_{4}} is present in the solution.

Note: We can use molarity for dilution of a solution as
M1V1=M2V2{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}
Where M1andM2{{M}_{1\,}}and\,{{M}_{2}}\, are initial and final molarity
V1andV2{{V}_{1}}\,and\,{{V}_{2}} are initial and final volume respectively.
Always remember to convert the volume into liters while solving the question related to the molarity. Always express the answer as per the rules of significant numbers and approximation for better calculations.