Answer:
The answer is "[tex]\bold{1.39 \ mg}[/tex]".
Explanation:
Please find the complete question in the attached file.
Total Moles of [tex]Mg_2^{+}[/tex] = moles of [tex]MgSO_4[/tex]
[tex]= \frac{mass}{molar \ mass \ of MgSO_4}\\\\= \frac{0.450}{120.37}\\\\ = 0.0037385 \ mol[/tex]
[tex]Mg_2^{+} + EDTA4^{-} \longrightarrow Mg(EDTA)2^{-}[/tex]
EDTA mol in 37.6 mL of solution = 50.0 mL of [tex]Mg2^{+}[/tex]
[tex]= \frac{50.0}{500} \times[/tex] total moles of [tex]Mg2^{+}[/tex]
[tex]= \frac{50.0}{500} \times 0.0037385\\\\= 3.7385 \times 10^{(-4)} \ mol\\[/tex]
[tex]Ca2^{+} + EDTA4^{-} \longrightarrow Ca(EDTA)2^{-}[/tex]
[tex]CaCO_3 Moles = Ca2^{+} Moles =[/tex] EDTA moles in a solution of 1.40 mL
[tex]= \frac{1.40}{37.6} \times 37.6 \ mL[/tex] the solution of EDTA moles.
[tex]= \frac{1.40}{37.6} \times 3.7385 \times 10^{(-4)} \\\\= 1.392 \times 10^{(-5)}\ mol\\\\[/tex]
Mass of [tex]CaCO_3 =[/tex]mole[tex]\times[/tex] the molar mass of [tex]CaCO_3[/tex]
[tex]= 1.392 \times 10^{(-5)} \times 100.09\\\\= 0.00139 \ g\\\\ = 1.39 \ mg[/tex]
