Calculating Molarity, Solving for Moles & Grams, 4 Practice Examples

This video tutorial shows how to solve for the molarity of a solution. Molarity is calculated by dividing the moles of solute by liters of solution. You will also be shown how to determine the amount of mass or volume, given the molarity. Molarity is another way to express the concentration of a solution. In everyday life we call it concentration. In chemistry, we call that MOLARITY!

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Thanks for the kind words, sometimes chem is a foreign language!

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[@sciencepost]

Thanks for the feedback. I'm glad the video worked you you. I did the best that I could when I made them. Keep up the great effort on your part!

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[@sciencepost]

Thanks for taking the time to work out the problem and leaving a comment.

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I didn't understand any bit of this in class, but your video makes it seem so easy! Thank you :D

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Alright! You made it so easy. I just realized that I am pretty good at dimensional analysis 👏

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Thank u so much ... understand nicely

[@rickeishasteele4083]

best video as help me so much of the confusion of molarity problem I thank you so much for making this video,,,,,,, hope I pass my test this Friday

[@sciencepost]

rickeisha steele Thanks for the feedback bro. Now that you understand it, just put the time into practicing it and you'll do great. Best wishes, let me know how it goes.

[@sciencepost]

Stoked that it worked for you!

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I FINALLY GET THIS THANK YOU

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learned more from this video that I spent 500 dollars on a class. fuck college.

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+082256694 Seriously! My lecture professor only reads the powerpoint and doesn't go in depth into the mathematical problems.

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Thanks!

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great video it really helped

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Awesomeness!

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[@danosthanos5766]

What if all you have is mL? And you're looking for moles? Do you get the molarity from the balance equation?

[@voicestar1084]

Thank s

[@smeighan216]

did you move the decial in the 228 to .228 because of the sig figs?

[@reymonroy]

Thanks

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LOVE YOUR VIDEOS !!! Im wondering if you can help me out with this problem ! 2 H2(g) + O2(g) → 2 H2O(l) Suppose we carry out the above reaction with 30.0 g of O2 and 4.00 g of H2. (a) Which reactant is the limiting reactant? (b) How much of the other reactant remains at the end of the reaction? (c) What mass of water is produced? In part a - I found that the limiting reactant is oxygen and hydrogen is in excess, can you help me with the other two please?

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[@cr125153]

In the last problem i tried it myself before you would do it and i endid up doing it differently and i dont know if it is right or wrong ? here is what i did. i tranformed 150ml into 0.15l i multiplied 58.5(NaCL) times 3mol =175.5 and then i divided 175.5 and 0.15= 26.32g

[@angthomas7241]

how do i work out the volume if not given

[@mrorange3529]

When do we use 1000/25 and when do we use 25/1000 cm3

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my favorite part was when he said: " now we're all set dudes." 5:15

[@sciencepost]

Let me know how the test goes! Best wishes

[@thericate]

(you dont have to solve it, just my question in the bottom) So what if it asks the molarity of Na+ in the solution when .20 mol of Na2SO4 is dissolved in water to produce 100 mL of solution. Do I have to multiply by 2 in order to get the molarity of Na+*

[@sciencepost]

thericate Bingo. Bear in mind that the molarity expressed is for the ionic compound. When it dissociates, you'll have 2x the Na+ ions. In the end, you'll have to divide by the volume to get the molarity.

[@ramona283]

Isn’t 5L of water the liters of solvent and NOT the liters of solution? Im confused

[@masilelaluckyg]

The molar masses are not correct that are given from the video try to check them again

[@reyquek3436]

How do you get 111g/mol?? If it's not given?

[@UpwardShotsStudios]

+RaxrQie Beetle He took the Atomic Masses of Ca and Cl2 and added them (40.1+70.8~111)

[@cr125153]

Sorry i didnt divided 175.5 and 0.15 i multiplied them.

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[@sciencepost]

Thanks so much! Keep being awesome!

[@mayaghosh6593]

First question: instead of writing 5.0 L of water, write 5.0 L of SOLUTION.

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Hey man that is soooo cool and all, but who asked

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Horrible explanation

[@lcdp9459]

Excellent! Thx, babe! 💛