What is the concentration of each of the following solutions?
(b)...

Question

What is the concentration of each of the following solutions?
(b) The mole fraction of each component of a solution pre- pared by dissolving 2.25 g of nicotine, C10H14N2, in
80.0 g of CH2Cl2

Accepted Answer

well, everyone's on this video of being asked what is the mole fraction of the components of a solution that's made by 2.15 g of Ben's OIC acid dissolved in 50 g of acetone were also given the molar masses of each component here. So first we can go ahead and start by calculating for the moles of each component. Starting off with our benzoate acid. So given the mass year of 2.15 g of Ben's weak acid, we can go ahead and use the molar mass to convert our grams into moles. So for every 122.12 g we have one mole of Ben's OIC acid. So you see here that the units of grams will cancel. And finally playing this into my calculator, I get the value of 0. units being moles. And same thing with our acetone. So given that we have 50 g of acetone again, we'll use the molar mass here to convert our grams into moles. So for every 58.8 g we have one mole of a stone. You see here that the gram unit will cancel. Now putting this into my calculator, I get the value of 0. moles. And with this information for the molds of each component, we can calculate for the mole fraction of each. So starting off with the mole fraction of Ben's weak acid. This is equal to well the moles of Arbenz folic acid on our numerator numerator. So that 0.176056 moles. And the denominator has to be the total amount of moles. So it's going to add the moles of both, that 0.176056 plus 0.8608815. And then once you get that, we get the value of 0.02 mol being the mole fraction of Urban's OIC acid, calculating for the mole fraction of acetone. Again, As our numerator, we need the moles of our stone. So that's 0. 08815 moles. Denominator will be exact same. So that's 0.176056 plus 0.8608815. So once you put that into a calculator, you get the value of 0.98. So, these values here are going to be my final answer for my problem. All right. Thank you all so much for watching

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