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Chapter 4: Problem 35

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: 100.0 \(\mathrm{mL}\) of $0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{mL}\( of \)0.60 \mathrm{M} \mathrm{MgCl}_{2},$ or 200.0 \(\mathrm{mL}\) of 0.40\(M \mathrm{NaCl} ?\)

Short Answer

Expert verified
The 100.0 mL of 0.30 M AlCl3 solution contains the largest number of moles of Cl- ions, with 0.09 mol of Cl-.

Step by step solution

01

Calculate the moles of AlCl3

To find the moles of AlCl3 in the first solution, we multiply the volume of the solution (100.0 mL) by its concentration (0.30 M): moles of AlCl3 = Volume * Concentration = 100.0 mL * 0.30 mol/L Since 1 L = 1000 mL, we can convert mL to L: moles of AlCl3 = 0.100 L * 0.30 mol/L = 0.03 mol
02

Calculate the moles of MgCl2

To find the moles of MgCl2 in the second solution, we multiply the volume of the solution (50.0 mL) by its concentration (0.60 M): moles of MgCl2 = Volume * Concentration = 50.0 mL * 0.60 mol/L Converting mL to L: moles of MgCl2 = 0.050 L * 0.60 mol/L = 0.03 mol
03

Calculate the moles of NaCl

To find the moles of NaCl in the third solution, we multiply the volume of the solution (200.0 mL) by its concentration (0.40 M): moles of NaCl = Volume * Concentration = 200.0 mL * 0.40 mol/L Converting mL to L: moles of NaCl = 0.200 L * 0.40 mol/L = 0.08 mol #Step 2: Calculate the moles of chloride ions in each solution#
04

Calculate the moles of Cl- from AlCl3

One mole of AlCl3 dissociates into 3 moles of Cl- ions when dissolved in water. Therefore, to find the moles of Cl- ions from AlCl3, we multiply the moles of AlCl3 by 3: moles of Cl- from AlCl3 = 3 * moles of AlCl3 = 3 * 0.03 mol = 0.09 mol
05

Calculate the moles of Cl- from MgCl2

One mole of MgCl2 dissociates into 2 moles of Cl- ions when dissolved in water. Therefore, to find the moles of Cl- ions from MgCl2, we multiply the moles of MgCl2 by 2: moles of Cl- from MgCl2 = 2 * moles of MgCl2 = 2 * 0.03 mol = 0.06 mol
06

Calculate the moles of Cl- from NaCl

One mole of NaCl dissociates into 1 mole of Cl- ions when dissolved in water. Therefore, the moles of Cl- ions from NaCl are equal to the moles of NaCl: moles of Cl- from NaCl = moles of NaCl = 0.08 mol #Step 3: Compare the moles of chloride ions to find the largest#
07

Determine the largest number of moles of Cl-

Now compare the moles of Cl- ions in each solution: 0.09 mol (from AlCl3) > 0.06 mol (from MgCl2) and 0.09 mol (from AlCl3) > 0.08 mol (from NaCl) Since the moles of Cl- ions from AlCl3 (0.09 mol) are greater than those from MgCl2 and NaCl, the 100.0 mL of 0.30 M AlCl3 solution contains the largest number of moles of Cl- ions.

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Most popular questions from this chapter

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 \(\mathrm{mL}\) of 0.213 \(\mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required 13.21 \(\mathrm{mL}\) of 0.103 \(\mathrm{M}\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) . Sulfuric acid has two acidic hydrogens.

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