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# How much $Ca{\left( {N{O_3}} \right)_2}$, in $mg$, must be in $50\,ml$ of a solution with $2.35\,ppm$ of $Ca$?A. 0.1175B. 770.8C. 4.7D. 0.48

Last updated date: 11th Sep 2024
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Hint: A solution is marked by the presence of two parts. One part is the solute, and the other part is called the solvent. The solute is the substance that is dissolved in the solution. The solvent is the part where the solute is allowed to dissolve.

Given,
Volume of the solution is =$50\,ml$
Amount of $Ca$ present = $2.35\,ppm$
Now we will have to calculate the amount of $Ca{\left( {N{O_3}} \right)_2}$, in $mg$ present.
We know that, for $2.35\,ppm$ of $Ca$,
$\;{10^{ - 6}}{\text{ml}} = {\text{ }}2.35{\text{ mg}}\,{\text{of}}\,{\text{Ca}}$
So, for $50\,ml$,
$2.35\,{\text{ppm = }}\dfrac{{{\text{mass}}\,{\text{of}}\,{\text{solute}}\,\,{\text{in}}\,\,{\text{mg}}}}{{{\text{volume}}\,{\text{of}}\,{\text{solution}}\,\,{\text{in}}\,\,{\text{litres}}}} \\ \therefore 2.35\,{\text{ppm}} = \,\dfrac{{{\text{mass}}\,{\text{of}}\,{\text{solute}}\,\,{\text{in}}\,\,{\text{mg}}}}{{50 \times {{10}^{ - 6}}}} \\ \Rightarrow {\text{mass}}\,{\text{of}}\,{\text{solute}}\,\,{\text{in}}\,\,{\text{mg = }}\,2.35 \times 50 \times {10^{ - 6}} \\ = \,117.5 \times {10^{ - 6}} \\$
Now, we will have to calculate the molar mass of $Ca{\left( {N{O_3}} \right)_2}$
Molar mass of $Ca{\left( {N{O_3}} \right)_2}$=$40 + 28 + 96 = 164\,{\text{grams}}$
Now, we will calculate the mass of $Ca{\left( {N{O_3}} \right)_2}$
$164\,{\text{g}}\,{\text{of}}\,{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}} = 40\,{\text{g}}\,{\text{of}}\,{\text{Ca}} \\ = \,117.5 \times {10^{ - 6}} \times 164\,{\text{g}}\,{\text{of}}\,{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}} \\ = \,481.75\,{\text{g}} \\ = 481.75 \times {10^{ - 3}}\,{\text{mg}} \\ = \,0.48\, \\$
Therefore, out of the given four options, (D) is the correct option. A, B and C are incorrect options.

${\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}} + {\text{2HN}}{{\text{O}}_{\text{3}}} \to {\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}} + {\text{C}}{{\text{O}}_{\text{2}}} + {{\text{H}}_{\text{2}}}{\text{O}}$
${\text{C}}{{\text{a}}_{\text{3}}}{\left( {{\text{P}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} + 6{\text{HN}}{{\text{O}}_{\text{3}}} + 1{\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \to 2{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}} + 3{\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}} + 12{{\text{H}}_{\text{2}}}{\text{O}}$
Note:Calcium nitrate has the chemical formula ${\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$. This compound is also called Norges Saltpeter. The structure of this compound is written as follows.