A typical oral rehydration solution (ORS) for infants contains 90...

Question

A typical oral rehydration solution (ORS) for infants contains 90 mEq/L Na⁺ , 20 mEq/L K⁺ , 110 mEq/L Cl⁻ and 2.0% (m/v) glucose (MW = 180g/mol)a. Calculate the concentration of each ORS component in units of molarity.

Accepted Answer

Welcome back everyone. A certain sports drink has the following ingredients. 1.10 milli equivalent per liter of calcium Calion, 0.494 milli equivalents per liter of magnesium two plus Calion 22.6 milli equivalents per liter of chloride anion and 0.42% mass by volume fructose. What is the concentration of each ingredient in the molar or in molar which is in milli molar? And note that the molecular weights of fructose is 180.16 g per mole. Let's recall that molar which is a capital m is expressed in terms of moles of solute per liter of solution. And because we are determining the polarity in milli molar, we would instead be interpreting minimal of each solute per liter of solution. And that would be our expression of milli molar. So let's begin by taking our given 1.10 milli equivalents per liter of calcium cad. And we're going to multiply by a conversion factor to go from Mila equivalent in the denominator to then millimoles of charge in the num reader. And so based on our textbooks, we would recall that one mili equivalent is equivalent to one millimoles of charge for an ion. So we can then cancel out mili equivalents in our numerator and Zeno and we have millimoles per liter. However, we specifically need to also incorporate a second conversion factor where we would go from millimoles of charge in the denominator to millimoles of our ion calcium to plus in the numerator. And so we would recall that from our ion charge of calcium two plus, we can interpret that for one millimoles of calcium Calion in the numerator, we've got two millimoles of charge. And so we can then cancel out the units millimoles of charge. And we are officially left with millimoles of calcium Calion per liter. And so within the sports drink, we can interpret that we've got 0. mili molar of calcium two plus caton as the concentration in the sports drink. Now, let's move forward to our second ion where we are given from the prompt 0.494 mila equivalences per liter of magnesium two plus. So we would multiply to go from again, one milli equivalent in the denominator to one milli mole of charge in the numerator. So canceling out millet equivalents, we've got millimoles of charge per liter and we are then going to multiply to go from millimoles of charge in the denominator to millimoles of magnesium two plus in the numerator. And just as before, based on that plus two charge we've got two millimoles of charge for one millimoles of magnesium two plus C. So canceling out millimoles of charge, we are left again with millimoles of magnesium ion now per liter. And this will simplify so that we find our concentration equal to 0.247 milli molar of magnesium two plus as the concentration in the sports drink. So now let's move on to our third ion mentioned where we are given 22 0.6 milli equivalents per liter of chloride anions. And we will then of course multiply to go from one milli equivalent in the denominator, which has an equivalent of one milli mole of charge in the numerator. So canceling out miller equivalents, we are left with minimal of charge per liter and we will then multiply to go from millimoles of charge in the denominator to millimoles of chloride ions in the numerator where from our one minus charge for chloride, we can say we have one milli mole of charge in the denominator equivalent to one millimoles of chloride ion. And so canceling out millimoles of charge, we are left with millimoles of chloride per liter. And within the sports drink, we would find a concentration of chloride anions equal to the value of 22.6 milli molar and sorry, this is milli molar. So we need a small M and then a capital M of chloride and ions. And so now we then would have our fourth ion mentioned. And rather this is a sugar mentioned which is a concentration of 0.402 or we can say composition of 0.402 grams per 100 mL of solution. And so we have grams of fructose in the numerator of course. And so we would multiply first to cancel out the unit of milliliter. So we have milliliters in the in the numerator and liters in the denominator. And recall that our prefix Milly tells us we have an equivalence of 10 to the negative third of our base unit liter. So canceling out milliliters, we have grams of fructose per liter of solution and we will then introduce moles as our next conversion factor. So we will multiply to go from grams in the denominator to moles in the numerator where we would recall the molar mass of fructose. So for one mole of fructose in our periodic table, we can count a molar mass of 180. I'm sorry, that's 180 0. g. So that's grams per mole. And so canceling out now, grams of fructose, we have moles per liter, we need to keep going so that we can then get to millimoles per liter. So we will make our next conversion factor that we multiply by be going from moles in the denominator to then millimoles in the numerator. So we would recall that our prefix Milli tells us that we have an equivalence of 10 to the negative third of our base unit mole. So canceling out moles, we've got millimoles in the numerator and this is per liter. And so we would simplify so that we find the concentration of fructose equal to a value of 22.3 milli molar of fructose as the concentration of the sugar in the sports drink. So our final answers were will be sum summarized as a total of four final answers where our concentration of calcium in the sports drink is 40.5 50 milli molar. Then our concentration of magnesium two plus in the sports drink is 20. milli molar. Our concentration of chloride in the sports drink is going to be 22. milli molar. And our fourth answer is our concentration of fructose in the sports drink which is 22.3 milli molar. I hope that this made sense and let us know if you have any questions.

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A typical oral rehydration solution (ORS) for infants contains 90 mEq/L Na⁺ , 20 mEq/L K⁺ , 110 mEq/L Cl⁻ and 2.0% (m/v) glucose (MW = 180g/mol)a. Calculate the concentration of each ORS component in units of molarity.

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