Problem: Which one of the following is not true concerning 2.00 L of 0.100 M solution of Ca3(PO4)2?A) This solution contains 0.200 mol of Ca3(PO4)2. B) This solution contains 0.800 mol of oxygen atoms. C) 1.00 L of this solution is required to furnish 0.300 mol of Ca2+ ions. D) There are 6.02 x 1022 phosphorus atoms in 500.0 mL of this solution. E) This solution contains 0.600 mol of Ca2+.

$\boxed{\mathbf{Molarity}\mathbf{ }\mathbf{\text{(M)}}\mathbf{=}\frac{\mathbf{moles}\mathbf{ }\mathbf{solute}}{\mathbf{L}\mathbf{ }\mathbf{solution}}}$

solute → Ca₃(PO₄)₂ 

Ca₃(PO₄)₂ → 3 Ca²⁺ + 2 PO₄^3-

A) This solution contains 0.200 mol of Ca₃(PO₄)₂. 

$\mathbf{moles}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{L}}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}}{\cancel{\mathbf{L}}}$

moles Ca₃(PO₄)₂ = 0.200 mol Ca₃(PO₄)₂

The statement is TRUE.

B) This solution contains 0.800 mol of oxygen atoms. 

$\mathbf{moles}\mathbf{ }\mathbf{O}\mathbf{ }\mathbf{atoms}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{L}}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}}{\cancel{\mathbf{L}}}\mathbf{\times }\frac{\mathbf{8}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{O}\mathbf{ }\mathbf{atoms}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}}$

moles oxygen atoms = 1.600 mol O atoms

The statement is FALSE.

C) 1.00 L of this solution is required to furnish 0.300 mol of Ca²⁺ ions. 

$\mathbf{moles}\mathbf{ }{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{L}}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}}{\cancel{\mathbf{L}}}\mathbf{\times }\frac{\mathbf{3}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}}}$

moles Ca²⁺ = 0.300 mol Ca²⁺

The statement is TRUE.

D) There are 6.02 x 10²² phosphorus atoms in 500.0 mL of this solution.