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Chemistry 22 Online
OpenStudy (anonymous):

If you had only 19g of KOH remaining in a bottle, how many milliliters of 9.5%(w/v) solution could you prepare? *Answer: 200 mL How many milliliters of 0.21M solution could you prepare? *I converted Moles to g and got 11.78g, which I then used the same formula I used above and got 124 mL but it says it's Wrong!! ): Can someone tell me what Im doing wrong!!): Pleaseeeeeeeeeee.. I just wanna go to bed ):

OpenStudy (chmvijay):

9.5 W/V mean 9.5 gram is there in 100ml of water ok then 19 gram will be there in = 100*19/9.5

OpenStudy (anonymous):

Thanks for the reply!(: Yes, I got that exact Answer!! I'm only confused about the second question where it asks --- How many mL of 0.21M could you prepare? I did all the work but it says my answer "124 mL's is wrong".):

OpenStudy (chmvijay):

KOH molar mass?

OpenStudy (anonymous):

KOH molar mass is 39.10 +16+1.01=56.11

OpenStudy (chmvijay):

56.11 g in 1000 ml gives 1M 19 g in 200 ml gives how much molar ?

OpenStudy (anonymous):

Don't I have to convert Moles to grams by multiplying (56.11g/M)(0.21M) though?

OpenStudy (chmvijay):

u first find molarity of ur solution :)

OpenStudy (anonymous):

Oh Okay so I got..0.05611.. Then for 19g in 200ml I got 0.095.. So now do I divide 0.056 into 0.095??

OpenStudy (chmvijay):

0.095*200=0.21*V

OpenStudy (anonymous):

I'm sorry I don't understand that.): Why would you multiply it by 200??

OpenStudy (chmvijay):

bcoz u have 200 ml of 0.095 M solution from that u have to prepare the 0.21 ml solution :) the question is how much u have to prepare ?

OpenStudy (anonymous):

Doesn't it say to find out how many mL? 200mL was the mL for the first question, on the second one it's different... I'm sorry I'm taking forever to understandd this..

OpenStudy (chmvijay):

here also ur finding the volume itself it will come around 90.4 ml is that ur answer

OpenStudy (anonymous):

No, it says that's wrong.. and I'm sorry but I'm really not understanding what you're saying. Why are we finding the volume?? I Just need to know the steps on how to do it...

OpenStudy (anonymous):

It just seems like it should be a lot more simple than this. I don't know if I'm over thinking this whole problem or what I'm doing wrong but I'm just getting more and more confused..

OpenStudy (zpupster):

molec. mass of KOH = 56.11 g/mol 19 g = 19/56.11 moles = 0.339 moles KOH. in a 0.21 M solution, there are 0.21 moles in 1000 mL 1 mole is in 1000/0.21 mL and 0.339 moles are in 0.339*1000/0.21mL = 1614 mL <--- ans.

OpenStudy (zpupster):

is that your answer??

OpenStudy (anonymous):

Thanks for replying! I'll check!

OpenStudy (anonymous):

I need your address... TO SEND YOU FLOWERS!!! Hahaha.. YOU ARE AMAZING, THANK YOU! I WAS SO FRUSTRATED I ALMOST CRIED! THANK YOU SO MUCH!!!!!(:

OpenStudy (zpupster):

your welcome!!

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