Molarity, Solution Stoichiometry and Dilution Problem
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- Опубликовано: 25 июл 2024
- This example shows three different types of ways a solution stoichiometry question can be asked, using molarity, stoichiometry and dilutions. I walk you through how to solve each part while showing you all the steps of course!
Chemistry Question:
According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density of 1.18g/mL. What is the molarity of concentrated HCl? What volume of it would you need to prepare 283mL of 2.20M HCl? What mass of sodium bicarbonate would be needed to neutralize the spill if a bottle containing 1.75L of concentrated HCl dropped on a lab floor and broke open?
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For part b. Doesn't V2 have to be in Liters?
very good tech
can you do particles to mole and mole to particles
We can also calculate Molarity for question.a)with this formula and gives an answer 11.6 M
M=10×density×% by weight of solute/molecular mass of solute
M=10×1.18×36/36.45
That gives 11.6M molarity.
Why 3 sig figs in the answer ?
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how did we get 11.7 M HCl, what did I have to add/sub./divide/multiply to get that?
Hey Melissa at 0:53, where did you get the 100 g HCL solution from? How come you didn't put it over 1.18 g/mol
Hopefully, I can answer that question for you although I'm not Melissa. In Chemistry, we typically assume that we have 100g of a certain solution unless otherwise specified. Now the 1.18g/mol has a different role in this problem. Because it's g/mol, you use that to convert the g of HCL to g of HCL divided by 1 ml of HCL solution. I hope that helped you, even a little bit.
Hey mellee. The mass of Hcl is given in percentage form and it must be out of 100%.
Melissa, I’m so confused bcz my teacher never used conversion factor 😭😭😭😭 my finals coming soon n I’m so stressed
In the first question Why our answer has 3 sig figs while the least is 2 ?
Do you have a version of this video without the music? It's hard to follow with the music going.
Nope
How do you get from 1.18 g HCL to 0.4258 g HCL, you say to divide by the g of HCL by Ml of HCL solution, but the initial thing doesn’t give the ml of HCL solution. Could somebody please help?
Because, 36g HCL / 1 ml HCL solution doesn’t equal to 0.4248 g HCL / 1 ml HCL solution
Can I ask you a question?
Melissa: We're asked 3 different questions here.
Me not seeing any questions: 🥺😫
Oh wait. I just realized the questions came after. I thought she got all those questions from the word problem alone 😂 That's how stressed I am.
where is the number 0.4248 g HCL coming from?
It's in part a at 1:17 , it's just changing grams of HCL solution to grams of HCl.
why did it became 0.4248 g HCl in ? hopefully you see this
in the beginning,, why 100g HCL solution?
I think it's because HCl is 36% by mass of the HCl solution and this is the the same as saying 36/100 or "36 g HCl OVER 100 g HCl solution" I had the same question and this is where I think the 100 came from. Hopefully someone can confirm this
I dont get why you do 36g HCL/100g HCL * 1.18/
We are given 36% by mass, the formula for percent by mass is (grams of solute/ grams of solution) x 100, this is where the 36 g HCl/ 100 g HCl solution comes from. Next we are looking for the molarity of HCl, the units for molarity are the mol of HCl solute/ L HCl solution. We need to get rid of the grams of HCl solution since that is not part of the molarity formula. This is why I needed to cancel out the grams of HCl solution units by multiplying by the density as shown at 1:29 This allows me to get to g HCl solute/ mL HCl solution. Now we must get to the units for molarity are the mol of HCl solute/ L HCl solution. This is why I converted the g HCl solute to moles and the mL HCl solution to L. Which will give us our final answer of molarity of HCl .
can you show how to do grams of a product to prepare a solution
Hey Jacquelyn, did you have a specific question in mind?
The question is how many grams of Mg(OH) are needed to prepare 125ml pf 0.30M solution?
Ok so we will use the molarity as our conversion factor since M= mol/L this is really telling us 0.30 mol Mg(OH)2/ 1 L
Here's the plan:
mL to L to mol to g
We'll start with the mL and convert to L , then use the molarity to cancel out the L to get to moles, lastly convert the moles to grams using the molar mass.
Looks like this:
125 mL x (1 L/ 1000 mL) x ( 0.30 mol Mg(OH)2/ 1 L )x (58.32 g Mg(OH)2 / 1 mol Mg(OH)2)
Make sense?
yes. Thank you
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