Lactose, C12H22O11, is a naturally occurring sugar found in mamma...

Question

Lactose, C12H22O11, is a naturally occurring sugar found in mammalian milk. A 0.335 M solution of lactose in water has a density of 1.0432 g/L at 20 °C. What is the concentra- tion of this solution in the following units?
(b) Mass percent

Accepted Answer

Hello everyone today. We have the following problem. Carbon dioxide is a byproduct during the fermentation of sugars at 25 degrees Celsius. The density of 1.25 molar solution of carbon dioxide and water is 1.449 g per middle leader determine the concentration of the solution in mass percent. So the very first thing you wanna do is we want to make note of what the molar mass is of carbon dioxide. So we see we have one carbon And we have two oxygen's. We're going to multiply that by the molar mass that we see on the periodic table. So we know that carbon has a molar mass of 12.1 g per mole g to moles. And then oxygen has a molar mass of about 16 g per mole. And so when we have this up, this comes to 44 . g per mole next, we need to assume that we are using one leader of solution just to simplify things. And from that we can go ahead and find the mass of our solution And then it's gonna be found by taking 1000 ml and multiplying that by our density, Which is 1.449 g per one militia leader. When are units of middle leaders canceled out. We're left with 1,449 g. What we're going to do next is we're going to find the mass of our salute this time. However, we do already have our units of what we need. So we're gonna do is we're gonna take our 1.25 moles of our ceo to notice how the problem. It says 1.25 moller however polarities and units of moles per one liter. So we already assume that we're using one leader. So we just simply have 1.25 moles of carbon dioxide. So we're going to take those moles. We're going to multiply that by the molar mass of carbon dioxide which is 44 point oh one g of carbon dioxide over one mall of C. 02. When our moles of C. O. to cancel that, we're left with 55 .01 - five g of CO2. Now we can use our masses and find the mass percent our mass percent of C. 02. So we take our 55.1 to 5 g and divide that by r 1449 g. You multiply that by 100 since it is mass percent and we get 3.8% overall. I hope this helped. And until next time

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