A laboratory experiment requires 4.8 L of a 2.5 M solution of sulfuric acid (H2SO4), but the only available H2SO4 is a 6.0 M stock solution. How could you prepare the solution needed for the lab experiment? Show all the work used to find your answer. @Abhisar
@Nurali
@paki
@mayankdevnani can u plz help?
From this whole question,we can conclude that moles are same when we dilute and un-dilute acid.
correct or not ?
yes, correct!
\(\sf \huge M_1V_1=M_2V_2\)
so,we get \[ \bf Molarity~diluted \times volume~diluted=given~molarity \times given~volume\]
u just copied and pasted that from yahoo answers @Nurali
understood or not? @iamabarbiegirl
just abhisar write it in short form
oh ok
\(\sf 4.8\times 2.5=6\times V\) Now find the value of V
then plug values,we get \[\large \bf 2.5 \times 4.8=6.0 \times x(volume)\] now solve for x
LOL ! yes
we divide both sidea by 6? @Abhisar
yeah and then simplify it
\(\color{blue}{\text{Originally Posted by}}\) @Abhisar \(\sf \huge \frac{(4.8\times 2.5)}{6}=V\) Now find the value of V \(\color{blue}{\text{End of Quote}}\) yes..
v= 2?
yes...
Since we need 4.8 L. We will have to add (4.8-2=2.8 L) of water into it
y did u subtract 2 from 4.8?
Because we already have 2 L of 6 M solution. To make it 4.8 L of 2.5 M solution we will have to add 2.8 L of water.
Now after adding water we will get 2+2.8=4.8 L
ooohh okkk
Still confused ?
no i got it couldn't have said it better myself lool
Sure..be honest !
lol i am honest thats why i get all the stuff u explain to me ad am able to put in all in my own words with ur help
that's great :) \(\huge\sf\color{green}{\text{✌゚\(\ddot\smile\) ✌゚}}\)
haha thank you now i will put it in my own words and check it for me plz
sure !
does that look good? @Abhisar
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