The question is incomplete,complete question is:
An intravenous replacement solution contains 4.0 mEq/L of [tex]Ca^{2+}[/tex] ions. How many grams of
(Hint: 1 mole of [tex]Ca^{2+}[/tex] ions = 2 mEq of
Answer:
240.0 grams of [tex]Ca^{2+}[/tex] are in 3.0 L of the solution.
Explanation:
Concentration of calcium ions = 4.0 mEq/L
Volume of the solution = 3.0 L
Millimole equivalents of calcium ions = 4.0 mEq/L × 3.0 L = 12.0 mEq
1 mole of [tex]Ca^{2+}[/tex] ions = 2 mEq of
Moles of calcium ions in 12.0 mEq = [tex]\frac{12.0 }{2}mol=6.0 mol[/tex]
Mass of 6.0 moles of calcium = 6.0 mol × 40 g/mol = 240.0 g
240.0 grams of [tex]Ca^{2+}[/tex] are in 3.0 L of the solution.
The grams of Ca²⁺ in 3.0 L of the solution is 0.12024 grams
From the information given;
Now,
1 mEq of Ca²⁺ ions = atomic weight of Ca/valence electron
= 20.04 mg/2
= 10.02 mg
In 1 L solution that has 4.0 mEq, then there are:
= 4 × 10.02 mg of Ca²⁺ ions
= 40.08 mg of Ca²⁺ ions
Therefore, the grams of Ca²⁺ present in 3.0 L of the solution is:
= 3 × 40.08 mg of Ca²⁺
= 120.24 mg of Ca²⁺
= 0.12024 grams of Ca²⁺
Therefore, we can conclude that the grams of Ca²⁺ in 3.0 L of the solution is 0.12024 grams
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