A patient receives 3.2 L of intravenous (IV) glucose solution. If 100. mL of the solution contains 5.0 g of glucose (carbohydrate), how many kilocalories did the patient obtain from the glucose solution?

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Welcome back everyone. A type two diabetic patient was given 3.7 liters of an intravenous glucose solution. Calculate how many kilocalories can be obtained from the solution. If it contains 2.5 g of glucose or carbohydrate with an energy value equal to four point oh kilos per gram. And this is per 100 mL of solution. We're going to begin by finding our mass of glucose C six H 06 to find our mass of glucose. We're going to begin by taking the proportion of glucose solute that we have in our solution in grams, which is given as 2.5 g of glucose and this is per 100 mL of solution. Next, we're going to want to notice that we're given our total volume of our solution as 3.7 liters. However, we have inconsistent units of volume where in our ratio that we just wrote out between our cell ute glucose and the entire solution in our denominator, we have units of milliliters. And so we're going to need to multiply by a conversion factor to go from milliliters in the numerator to leaders in the denominator where we recall that our prefix Milly tells us that we have an equivalent of 10 to the negative third of our base unit leader in the denominator. So canceling out milliliters, we're then going to multiply by our given volume for the entire intravenous glucose solution, which has given us 3.7 liters. And now we can cancel out our unit of liters. We're left with units of grams for mass, which is what we would like. And in our next line, when we simplify, we'll find a mass of 92.5 g of glucose. Now, with our massive glucose to recall that from the prompt, we're asked to obtain the killer calories or to determine the killer calories that will be obtained from our solution given the proportion of our 2.5 g of glucose per 100 mL of solution. And so we're going to need to ultimately calculate our total energy gained from our glucose solution. We can find this by taking our mass of our nutrient within our solution. So within our glucose solution, and this is C six H oh two. Now, with the mass of our nutrient in our solution, we're then going to multiply by our energy value of our nutrient in our solution. So, plugging in what we know, we're going to recognize that the prompt tells us that glucose is a carbohydrate. And this would make sense because we recall that our structure of glucose being a ring is a monosaccharide. So ultimately, it's a single sugar unit consisting of carbon, hydrogen and oxygen. And because glucose is a monosaccharide, it's going to be the simplest form of a carbohydrate. So that is going to be our nutrient in our glucose solution. And we determined that our mass of our glucose is 92. g. So we're going to plug that mass in 92.5 g of carbohydrates. This is then multiplied by our energy value for a carbohydrate which we're told from our prompt is four point oh kilocalories per gram. Notice that we can now cancel our unit of grams leaving us with kill cows in our numerator as our final energy unit. And when we take the product between these two terms, we're gonna get our final total energy for the glucose solution equal to 370 kill the cows. And this would be our final answer as the energy and kilocalories that can be obtained from our intravenous glucose solution. So I hope that this made sense and let us know if you have any questions.

A patient receives 3.2 L of intravenous (IV) glucose solution. If 100. mL of the solution contains 5.0 g of glucose (carbohydrate), how many kilocalories did the patient obtain from the glucose solution?

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Key Concepts

Caloric Content of Carbohydrates

Volume to Mass Conversion

Unit Conversion