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smooch1020 smooch1020
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11 years ago
What is the concentration in molarity of a solution that contains 2.561 g (NH4)2SO4 in a total volume of 36.5 mL

The molar mass of (NH4)2SO4 is 132.15 g
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11 years ago
I could just tell you the answer but it's important you know how to find it.
The best way to solve these problems is to understand why the heck you would need to do it in the first place. So think of it as if you were actually in a lab and had to find the answer. A good story for this question would be that you are in lab and you find a chemical whose concentration has not been labeled but that you need it for a critical experiment. You see that the recipie for the solution has been written on it however so you know that you can calculate the concentration and find out whether it's too dilute or too concentrated for your experiments because either one could kill your cells in your experimental setup.

So:
Ok you know that molarity= moles per liter. So a 1M (NH4)2SO4 solution (or one molar solution) is equivalent to 132.15g of (NH4)2SO4 in one liter of water.
So that's great but the question says that there is 2.561g of the compound right in your solution right? So let's begin by setting up a ratio between the number of grams in one mole and the number of grams in the solution.  We'll use the common ground of molarity which will help us make a transition into 35ml after a couple steps.

132.15g/ 1 Liter = 2.561g/ x number of liters (simple ratio, this will tell us how many liters of water we would want to make a one molar solution with our starting 2.561 grams of (NH4)2SO4.

The answer here is x=0.0193 liters or 19.3mL. So we now know that if 2.561g of (NH4)2SO4 was in 19.3ml of water then that would give us a standard 1M solution. But the question says that our solution of interest contains 36.5ml of water. So immediately we know that because our solution has more water, it will be more dilute, which means our answer must be less than 1M (because low molarity means fewer moles in one liter of water and high molarity such as 2M would mean 2 moles of compound in one liter of water)

So now that we have the idea that our answer should be less than 1M lets find out exactly what it should be. Piece of cake:
It's all about proportional changes, what is the proportional change between 19.3ml and 35.6ml? Simply divide 19.3ml by 36.5ml (you do it in this order because you set 19.3 as the standard for 1M solution and you're measuring the change in proportion to the 1M standard). So you divide 19.3ml by 36.5ml and you get 0.52. 1
Ok so the proportional change is 0.52 which means that 19.3ml is almost half of 36.5ml. So going back to the problem, you know 19.3ml gives you a 1M concentration but according to the problem you have 36.5ml of water in the same 2.561 g of compound which means your compound has stayed the same but your water has doubled so the concentration should be halved (because the solution is becoming more dilute with the doubling of water right?). So if you multiply 1M by the proportional change in volume, which is 0.52, you get 0.52M. Voila! your answer is 0.52M.
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