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loyalgooner

You are given one litre 0.183 M HCl and one litre 0.381 M HCl. What is the maximum volume of 0.243 M HCl which you can make from these two solutions?
No water is added.

Plz help me with this one.

rooney

edit : didnt consider that no water is added . Gimme a minute.

Sahadevankazhani

no water added its not important.By the given Hci we can make a solution . Think it

rooney

I have solved it. Typing the solution. Hold on.

rooney

We want that final vol should be maximum. Thus , that vol. should be added first which given least no. of moles (and thus,more vol. can be considered ). So , we add entire 1L of 0.183M of HCl.

Now, we have 1L and 0.183moles of soln.

Let vol of 0.381M Hcl added to this existing soln. be x (in Litres)
So,
=> (Total no. of moles)/Total Vol = 0.243
=> (0.183 + 0.381x) / (1 + x) = 0.243
=>0.138x=0.060
x = 60/138 = 0.434 L = 434 mL

So, total solution volume = 1.434 L or 1434 mL

atavistic

Plz explain "We want that final vol should be maximum. Thus , that vol. should be added first which given least no. of moles (and thus,more vol. can be considered ). So , we add entire 1L of 0.183M of HCl. "

rooney

See the question states " What is the maximum volume of 0.243 M HCl which you can make from these two solutions? "

We want that the final mixture (which consists of certain volumes taken from both the given solutions) should have MAX volume BUT there is a limit to the no. of moles contained per unit volume. So, think about it, if you want the volume to be max, wouldnt you take the solution with LESS no. of moles per unit volume ?? Because, then you will have to take MORE vol. of solution to reach the desired molarity. So, we consider 1L of 0.183M solution to be taken completely as it will help us maximise the final volume of mixture made.

atavistic

Ye.Thnx for clearing the doubt.

loyalgooner

Thanks for the help. The answer is perfect.

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