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An analytical procedure requires a solution of chloride ions. how many grams of Cacl2 must be dissolved to make 2.15 L of 0.0520 M Cl-? [Atomic mass: Ca=40.1 g/mol; Cl=35.5 g/mol]

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The dissolution of CaCl2 in water has an equation written as:

CaCl2 = Ca2+ + 2Cl-

First, we need to calculate how many moles of Cl- we need from the volume and concentration to be used.

mol Cl- needed = 0.0520 mol / L Cl- (2.15 L) = 0.1118 mol Cl-

We can now find how much of the CaCl2 we need as follows:

0.1118 Cl- ( 1 mol CaCl2 / 2 mol Cl-) ( 111.1 g / 1 mol) = 6.21 g CaCl2

Answer:

Mass of CaCl₂ that must be dissolved is around 6.22 g

Explanation:

Calcium chloride dissociates to form Ca²⁺ and Cl⁻ ions as shown below:

[tex]CaCl_{2}\rightleftharpoons Ca^{2+}+2Cl^{-}[/tex]

Therefore, 1 mole of CaCl₂ corresponds to 1 mole of Ca²⁺ and 2 moles of Cl⁻

From the given information:

Moles of Cl-  = [tex]Volume*Molarity = 2.15L*0.0520moles/L = 0.112 moles[/tex]

Based on the stoichiometry:

[tex]Moles\ of\ CaCl2\ required = \frac{1}{2} * moles\ Cl- = \frac{1}{2}*0.112 = 0.056 moles[/tex]

[tex]Molar\ mass\ of\ CaCl2 = 40.1 + 2(35.5) = 111.1 g/mol\\\\Mass\ of\ CaCl2\ = moles*mol\ mass = 0.056\ moles*111.1\ g/mol = 6.22\ g[/tex]

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