Question

An analytical procedure requires a solution of chloride ions. How many grams of NaCl must

be dissolved to make 1.95 L of 0.0561 MCI?
A) 3.28 g
B) 0.595 g
C) 6.39 g
D) 1.64 g
E) 12.8 g

Answer 1

To prepare a 1.95 L solution of 0.0561 M Cl-, approximately 6.37 g of NaCl must be dissolved, since NaCl dissociates in a 1:1 ratio and has a molar mass of 58.44 g/mol. The most closely related option of choice is C) 6.39 g.

Step-by-step explanation:

The question asks us to calculate the mass of sodium chloride (NaCl) required to prepare a solution with a specific concentration of chloride ions (Cl-). Below are the calculation steps:

1. Determine the number of moles of Cl- needed for the solution using the given molarity (0.0561 M) and volume (1.95 L). The number of moles of Cl- is equal to the molarity times the volume.
2. NaCl dissociates into Na+ and Cl- ions in a 1:1 ratio, so the moles of NaCl needed will also be the same as the moles of Cl-.
3. Calculate the mass of NaCl needed by multiplying the moles of NaCl by its molar mass (58.44 g/mol).

To answer the question:

• Moles of Cl- required: 0.0561 M * 1.95 L = 0.109 mol Cl-
• Moles of NaCl required: 0.109 mol (since 1 mole of NaCl gives 1 mole of Cl-).
• Mass of NaCl required: 0.109 mol * 58.44 g/mol = 6.37 g of NaCl.

Therefore, we would need to dissolve approximately 6.37 g of NaCl to create a 1.95 L solution with a 0.0561 M concentration of chloride ions.