How many milliliters of a solution containing 2 mEq of potassium chloride (KCl) per milliliter should be used to obtain 2.98 g of potassium chloride. The answer is 20 mL but how?
To determine the number of milliliters of the 2 mEq/mL solution needed to obtain 2.98 g of potassium chloride (KCl), we need to use the following information:
1. The molecular weight of potassium chloride (KCl) is 74.55 g/mol.
2. 1 mole of any compound contains Avogadro's number (6.022 x 10^23) of particles.
3. The term "milliequivalent" (mEq) is often used in medical and pharmacological fields to express the amount of an ion in a solution, where 1 mEq represents the amount of that ion that can combine with or replace 1 millimole of hydrogen ions (H+).
With these details, we can follow these steps to find the answer:
Step 1: Calculate the number of moles of potassium chloride required:
Given mass of KCl = 2.98 g
Molecular weight of KCl = 74.55 g/mol
Moles of KCl = Mass / Molecular weight
= 2.98 g / 74.55 g/mol
= 0.04 mol
Step 2: Calculate the number of milliequivalents (mEq) of potassium chloride:
Since 1 mole of KCl contains 1 mole of potassium ions (K+) and 1 mole of chloride ions (Cl-),
the number of mEq of KCl is the same as the number of mEq of potassium ions.
Number of mEq of KCl = 0.04 mol
Step 3: Calculate the volume of the 2 mEq/mL solution needed:
Given concentration of the solution = 2 mEq/mL
We can use the equation: Concentration = Number of mEq / Volume
Volume = Number of mEq / Concentration
= 0.04 mol / 2 mEq/mL
= 0.02 mL
Therefore, to obtain 2.98 g of potassium chloride (KCl), we would need approximately 20 mL of the 2 mEq/mL solution.