An intravenous saline solution contains 147 mEq/L each of Na+ and CI-
How many moles of CI-
are in 0.700 L of the saline solution?
To find the number of moles of CI- in 0.700 L of the saline solution, we first need to find the number of moles of CI- in 1 L of the solution.
Since the solution contains 147 mEq/L of CI-, and 1 mole of CI- is equal to 1 Eq, then 147 mEq/L of CI- is equal to 147 moles/L.
Therefore, in 0.700 L of the saline solution, there would be 0.700 L * 147 moles/L = <<0.700*147=102.9>>102.9 moles of CI-.
To find the number of moles of Cl- in 0.700 L of the saline solution, we need to convert the given concentration from mEq/L to moles/L.
We know that the molar mass of Cl- is 35.45 g/mol.
First, let's convert the concentration of Cl- from mEq/L to moles/L.
1 mole of ions = 1 Eq of ions
Therefore, the mEq concentration of Cl- is the same as the moles of Cl-.
Therefore, the concentration of Cl- in moles/L is 147 mEq/L.
To convert mEq/L to moles/L, we need to divide the concentration by the valence of the ion. Since Cl- has a valence of 1-, we divide the concentration by 1.
So, the concentration of Cl- in moles/L is 147/1 = 147 moles/L.
Now, we can calculate the number of moles of Cl- in 0.700 L of the saline solution.
Number of moles = concentration × volume
Number of moles = 147 moles/L × 0.700 L
Number of moles = 102.9 moles
Therefore, there are 102.9 moles of Cl- in 0.700 L of the saline solution.
To find the number of moles of Cl- in 0.700 L of the saline solution, we first need to calculate the number of moles of Cl- in 1 L of the solution.
Given that the solution contains 147 mEq/L of Cl-, we can convert mEq to moles by dividing by the conversion factor, which is the molar mass of Cl-.
The molar mass of Cl- is approximately 35.45 g/mol (atomic mass of chlorine).
Therefore, 1 mole of Cl- is equal to 35.45 g.
To convert mEq to moles, we need to divide the mEq value by the equivalent weight of Cl-. The equivalent weight of Cl- is the molar mass divided by the charge of the Cl- ion, which is 1-.
So, the equivalent weight of Cl- is 35.45 g/mol / 1- = 35.45 g/mol.
Now, we can calculate the number of moles of Cl- in 1 L of the solution:
147 mEq/L * (1 mol/35.45 g) = 147/35.45 mol/L
Now, we multiply this value by 0.700 L, which is the volume of the saline solution given in the question:
(147/35.45 mol/L) * 0.700 L = 2.885 mol
Therefore, there are 2.885 moles of Cl- in 0.700 L of the intravenous saline solution.