An intravenous saline solution contains 154 mEq/L each of Na⁺ and Cl^–. How many moles each of Na⁺ and Cl^– are in 1.00 L of the saline solution?

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Welcome back everyone. A veterinary injection has the following contents. 130 milli equivalents per liter of sodium caton four milli equivalents per liter of potassium CAD 2.70 milli equivalents per liter of calcium Calion and 109 milli equivalents per liter of our chloride anion. As well as 28 milli equivalents per liter of our lactate three minus anion. We need to determine how many moles each of our pota of sodium potassium calcium and chloride ions as well as our lactate anion are in a liter of the injection solution. We're going to begin by focusing on our unit of equivalence. And we're going to recall that equivalence and let's write this out equivalence determine our number of moles of charge that an ion contributes to solution or two A solution. And equivalents are going to be a positive value. We're going to recall two conversion factors where our first describes that our moles of equivalent of an ion. So for example, one mole of an ion for an atom X where its ion charge is N plus would be equivalent to our moles of equivalent expressed by lower case N where for that ion X, its charge or its moles would be equal to its charge, which is also expressed as N. Here, we're also going to recall that one milli equivalent has an equivalence of 10 to the negative third equivalence. Let's begin with our first ion to determine our moles of our sodium ion specifically in one liter of injection solution. So we're going to begin with the contents of our sodium ion in the injection solution given as 130 milli equivalents per liter of solution. And we're going to begin by multiplying by a conversion factor to cancel out our units of mili equivalents in the denominator to get to equivalents in the numerator. As we noted above one milli equivalent is equal to 10 to the negative third of equivalents, meaning we can now cancel out milli equivalents and we're going to need to next cancel out liters and introduce units of moles. So let's begin by introducing our units of moles first where we would recognize that because we have a plus one charge for our sodium ca we can say that in our denominator, one equivalent of our sodium ion in our denominator is equal to one mole. I'm sorry, this is one mole of our sodium ion. So now we can cancel our units of equivalents. We have introduced moles and we also need to introduce units of liss. So this should be the volume of our injection solution which were given for each of our ions as one liter. So we're going to multiply by one liter. And this will then allow us to cancel our units of leaders and we are left again with moles as our final unit. So, simplifying in our calculators, we would find that our moles of sodium ion in one liter of injection solution is equal to about 1.30 times 10 to the negative first power moles of our sodium ion. This is going to be our first answer so far. And now we're going to need to find our moles of our next ion given which is our potassium ion again in one liter of solution. So beginning with our contents of potassium ion in our injection solution, which is given in our prompt as four point oh milli equivalent per liter. We're going to begin by canceling our units of milli equivalent by multiplying just as before by our conversion factor where one milli equivalent in the denominator is equivalent to 10 to the negative third of our base unit equivalent in the numerator. So canceling out milli equivalents, we are left with equivalence per liter. And next, we're going to cancel out our units of equivalent by introducing moles where based on our plus one ion charge for potassium Canion, we can say that for one mole of potassium caton in the num, we have one equivalent of potassium in the denominator. So canceling out units of equivalence. Now we're left with moles per liter and we need to cancel out liters by multiplying by our volume of our injection solution being one liter, which will now allow us to cancel out liters. And again, leaving us with moles of our potassium ion as our final unit. So in our next line, we would find that our moles of potassium ion in one liter of injection solution is equal to a value of 4.0 times 10 to the negative third power moles of potassium ion. And this is our second answer so far. And now let's move on to our third ion listed, which is our calcium Calion. So we need moles of our calcium two plus Calion in one liter of solution. So beginning with our contents of calcium ion in our injection solution from the prompt, we're given a value of 2.70 milli equivalents per liter. This is then multiplied to cancel our units of milli equivalence. Where we understand that in our denominator, one milli equivalent is equivalent to 10 to the negative third of our equivalent unit canceling out milli equivalents, we have equivalents per liter left. And now we're going to cancel out equivalents utilizing our ca charge, which is a two plus charge where we would understand that for one mole of our calcium two plus caton in the numerator, we have two equivalents of calcium. Again, that is based on our calcium two plus ca charge. Then we want to after canceling units of equi equivalence, cancel our unit of leaders by multiplying by our volume of our solution listed as one leader. So canceling out leaders were now left with units of moles of our calcium Canion. And in our next line, we will simplify so that we find our moles of calcium Calion in our solution equal to a value of 1.35 times 10 to the negative third power moles of calcium Calion in our injection solution. And this is our third final answer so far. Now let's move on to our next ion listed, which is our chloride anion. So we want to find moles of our chloride anion in one liter of solution. Next, in which from the prompt, we're given our contents of our chloride anion in the solution being milli equivalents per liter, multiplying two again cancel or units of mili equivalent. We have one milli equivalent in the denominator equal to 10 to the negative third of our base unit equivalent canceling out milli equivalents. We have equivalents per liter in which from our minus one anion charge of chloride, we would say that for one mole of our chloride anion in the numerator, we have an equivalence of one equivalent of chloride. So now canceling out our units of equivalents, we're left with moles per liter. And we can multiply it by our volume of our injection solution being one liter to cancel out now liters that leaves us with our final unit as moles of our chloride anion. And this will simplify so that we find our moles of chloride anion in our injection solution equal to a value of 1.9 times 10 to the negative first power moles of our chloride anion in our injection solution as our fourth answer so far. So now we're going to move on to our next ion which is our moles of our lactate anion C three H minus in one liter of solution. In which from the prompt, we're told that our contents of lactate in the injection solution is 28.0 milli equivalent per liter of solution canceling out milli equivalents by multiplying by one milli equivalent in the denominator equal to 10 to the negative third equivalence in the numerator, we will cancel out mill equivalents and be left with equivalent per liter. So now we will multiply by our next conversion factor to cancel out equivalents. And we would recall that for one equivalent of our lactate anion C three H 503, we have one equivalent of lactate per mole of our lactate anion. And this is a one minus anion which helped us define this conversion factor. So now canceling out our unit of equivalents, we're left with moles of lactate per liter. We're going to need to cancel out leaders by multiplying by our volume of our injection to solution being one liter. So now canceling out liters, we're finally left with moles as our final unit. And this will simplify so that we find that our moles of lactate in one liter of injection solution is equal to a value of two. I'm sorry, this is a two so of 2. times 10 to the negative second power moles of lactate C three H 5031 minus so of our lactate ion in our injection solution. And this would be our fifth final answer. And so in summary, we've determined that we have 1.30 times 10 to the negative first power moles of sodium caton in our injection solution. We determined that we have four point oh times 10 to the negative third moles of potassium ion in our injection solution. We have also 1.35 times 10 to the negative third moles of our calcium Canion in one liter of injection solution. And then we have one point oh nine times 10 to the negative first power moles of our chloride anion in one liter of our injection solution. And lastly, we have 2.80 times 10 to the negative second power moles of our lactate anion in our injection solution. So these are our final answers. I hope that this made sense. And let us know if you have any questions

An intravenous saline solution contains 154 mEq/L each of Na⁺ and Cl^–. How many moles each of Na⁺ and Cl^– are in 1.00 L of the saline solution?

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An intravenous saline solution contains 154 mEq/L each of Na+ and Cl–. How many moles each of Na+ and Cl– are in 1.00 L of the saline solution?

Verified video answer for a similar problem:

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An intravenous saline solution contains 154 mEq/L each of Na⁺ and Cl^–. How many moles each of Na⁺ and Cl^– are in 1.00 L of the saline solution?