Molarity, Solution Stoichiometry and Dilution Problem

[@Jennnvang]

I don't know what I would do without your videos! Thank you for all your help! I have shared all your videos with classmates as well.

[@shanehebel]

I cant believe I have been trying to figure this out for 3 days now and you just taught it in a way that I understand and can apply in less than 15 min.. Just amazing!!!

[@xxsheannaexx]

I took chemistry in highschool and totally failed all my exams. Im saying I got 3 right out of 50 questions haha Im taking chemistry in college now and I've been learning from all her videos! So far I've gotten B's on my exams! Thank you!!

[@melissamaribel]

Wow, that is an incredible improvement!! You should be so proud, I know I'm proud of you 😊

[@kaithlyn-in5es]

Thank u so much! I have an exam tomorrow and am freaking out... but thanks to the videos it gives a clearer understanding! I never knew these were so easy:)

[@WeAreObject]

Ayyyy I got an exam tomorrow, and this is on it. Thanks!!

[@trinasalas3117]

This was so stressful learning this today in class -.- but thank you so much for your help

[@mattbraga4033]

My head hurts

[@melissamaribel]

Hey Matt, let's fix that headache, what part isn't making sense?

[@corymelaniepearson8025]

@@melissamaribel all of it :)

[@shravansrinivasan4982]

Thank you for your clear explanation towards these concepts It is extremely helpful for my Ap Chemistry work and I got A- so far in semester 1 and now aims for A+ in semester 2. Which means I will get an A thanks to you. I hope this will help for me to get A 4 and 5 in the exam.

[@charistanbibimbia330]

Thank you for this exercise . You're a great teacher in chemistry .

[@fuzzers6233]

I’m figuring out how to do this while taking my test 👀

[@padmavathiganti8152]

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.

[@orhanavunduk4934]

Why 3 sig figs in the answer ?

[@anthonylebario4460]

Very helpful video! Please make more of these types of videos! It truly helps, especially for research lab.

[@nellyandende7722]

Thankyou my baby is enjoying your lessons

[@greatg5775]

YOU SHOULD DO a video on organic chemistry

[@melissamaribel]

Good News, I have a whole second channel on Organic Chemistry! Here's the link: kzbin.info/door/SzvQG5seY5dA59RIkZlicg

[@dianareyes807]

For part b. Doesn't V2 have to be in Liters?

[@AvaDevaud]

how did we get 11.7 M HCl, what did I have to add/sub./divide/multiply to get that?

[@bittypie]

Melissa: We're asked 3 different questions here. Me not seeing any questions: 🥺😫

[@bittypie]

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.

[@melleesamson1210]

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

[@meganhuang9062]

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.

[@mohamudomar5828]

Hey mellee. The mass of Hcl is given in percentage form and it must be out of 100%.

[@KKÍlYO]

not me watching this video on the same day as my anachem exam 💀 (THANK U SM)

[@hiwakarem7640]

very good tech

[@labibafarheen9563]

Melissa, I’m so confused bcz my teacher never used conversion factor 😭😭😭😭 my finals coming soon n I’m so stressed

[@marilynward8697]

Do you have a version of this video without the music? It's hard to follow with the music going.

[@dinoclasher4129]

Nope

[@orhanavunduk4934]

In the first question Why our answer has 3 sig figs while the least is 2 ?

[@bellering.930]

u da best fr!

[@melissamaribel]

Thank you so much Gabriel! 😊

[@kaylaaquilante9445]

Wow this is a good video

[@ktq2236]

can you do particles to mole and mole to particles

[@alyanakaufman1514]

why did it became 0.4248 g HCl in ? hopefully you see this

[@rorybaker5799]

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?

[@rorybaker5799]

Because, 36g HCL / 1 ml HCL solution doesn’t equal to 0.4248 g HCL / 1 ml HCL solution

[@ewwxaca]

I dont get why you do 36g HCL/100g HCL * 1.18/

[@melissamaribel]

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 .

[@nurhasanahibrahim7622]

Can I ask you a question?

[@isabellarodriguez3431]

in the beginning,, why 100g HCL solution?

[@alexism7294]

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

[@Sabrinaxoxo92]

I’ve been watching all of your videos, but this one is the most confusing for me ☹️

[@3456uify]

where is the number 0.4248 g HCL coming from?

[@melissamaribel]

It's in part a at 1:17 , it's just changing grams of HCL solution to grams of HCl.

[@paulvanlieshout369]

can you show how to do grams of a product to prepare a solution

[@melissamaribel]

Hey Jacquelyn, did you have a specific question in mind?

[@paulvanlieshout369]

The question is how many grams of Mg(OH) are needed to prepare 125ml pf 0.30M solution?

[@melissamaribel]

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?

[@paulvanlieshout369]

yes. Thank you

[@tejpalsahani9295]

Fantastic

[@darmadeit]

🐐🐐🐐

[@thelivingzombie8403]

Messi