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### Topic: Molar concentration  (Read 2391 times)

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#### zafar31

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##### Molar concentration
« on: May 22, 2015, 12:44:09 PM »
How many litres of (NH3) [NH][/3] (the volume is measured in standart conditions) must be dissolved in 200 g of a 10% solution NH4OH to obtain 15% solution of the ammonium hydroxide?

Please show elaborate solution! And sorry I could not use symbols properly!

#### zafar31

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##### Re: Molar concentration
« Reply #1 on: May 22, 2015, 01:01:55 PM »
answer must be 6.9 litres If you want to check

#### Arkcon

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##### Re: Molar concentration
« Reply #2 on: May 22, 2015, 02:49:16 PM »
I'd like to welcome you, zafar31: to the Chemical Forums.  According to the Forum Rules{click}, we want to see your attempt, so please click on that link, read the contents, and comply with our rules, so we can help you.

Can you begin to talk about something you recently learned that is pertinent this question? Can you start with the reaction of the reagent you're using, with water?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### zafar31

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##### Re: Molar concentration
« Reply #3 on: May 22, 2015, 03:06:31 PM »
I found one formula:
M1 * x1+ m2 * x2=x3(m1+m2)
m - mass of substance
x - concentration in percent
x3 - final percentage

So in this question m1 is unknown, as NH3 consists of only NH3 x1=100%, m2=200g x2=10%,
x3=15%. from this I GOT answer 11,7 g which means 0,69 moles of NH3 as 1 mole equals to 22,4
litres I assumed that answer will be {15.456} and it is incorrect.
I have solved many exercises in this way and they were correct.

#### zafar31

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##### Re: Molar concentration
« Reply #4 on: May 23, 2015, 03:29:25 AM »
If you can find mistake (or if you think that my way of solving is totally wrong) please let me know what should I change!!!

#### Borek

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##### Re: Molar concentration
« Reply #5 on: May 23, 2015, 06:36:35 AM »
Around 15.5 L of ammonia looks OK to me.
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#### mjc123

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##### Re: Molar concentration
« Reply #6 on: May 23, 2015, 06:55:13 PM »
I agree with 6.9 L; I think you are ignoring the difference between ammonia and ammonium hydroxide. To get a solution 15% in NH4OH, you add x g NH3 and get (nominally) x*35/17 g NH4OH in solution, so that
20 + x*35/17 = 0.15*(200 + x)

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