Question

How many grams of $CuS{{O}_{4}}$ would be dissolved in $5L$ of $0.5M$ $CuS{{O}_{4}}$ solution?

Answer 1

Hint: 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= $0.5M$
And volume of solution= $5L$
Also, from the definition on molarity we know that
$Molarity\,\left( M \right)=\dfrac{Number\,of\,moles\,of\,solute}{volume\,of\,solution\,\left( in\,L \right)}$
From the above formulae, we can calculate the number of moles as shown below
$Number\,of\,moles\,=\,Molarity\left( M \right)\,\times \,Volume\,of\,solution\,$
Therefore, we can say that
Number of moles of $CuS{{O}_{4}}$ = $0.5\times 5=2.5$moles
Now we know that $number\,of\,moles\,=\,\dfrac{given\,mass\,of\,compound}{molar\,mass}$
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
$(mass\,of\,CuS{{O}_{4}})=\,number\,of\,moles\,of\,CuS{{O}_{4}}\,\times \,molar\,mass\,of\,CuS{{O}_{4}}$
Now, firstly let us find out the molar mass of $CuS{{O}_{4}}$
We know that molar mass of Copper= $63.5 g/mol $
Molar mass of Sulphur = $32g/mol $
Molar mass of oxygen=$16 g/mol $
Therefore, Molar mass of $CuS{{O}_{4}}$= $63.5+32+64=159.6\,or\,160$g/mol
$\therefore mass\,of\,CuS{{O}_{4}}=\,\,2.5\,\times \,160=\,400g$
Hence, we can say that $160$ gram of $CuS{{O}_{4}}$ is present in the solution.

Note: We can use molarity for dilution of a solution as
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
Where ${{M}_{1\,}}and\,{{M}_{2}}\,$ are initial and final molarity
${{V}_{1}}\,and\,{{V}_{2}}$ 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.