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Chemistry Question (2 Viewers)

Gruma

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Could someone solve this question as I think the answer in the textbook is wrong.

30ml of 0.374 mol L -1 sulphuric acid is diluted so the concentration becomes 0.110 mol L-1. The volume of water that must be added is:
 

wogboy

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No of moles of sulphuric acid = concentration * volume
= 0.374 * 0.03 mol
= 0.01122 mol

Final Concentration = 0.110 mol/L
No of moles of sulphuric acid = 0.01122 mol

so total volume of water = (no of moles) / (concentration)
= 0.01122 / 0.11 L
= 0.102 L

so amount of water added = 0.102 - 0.03 L
= 0.072 L

So I guess it should be 0.072 L or 72 mL of water added.
 
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kini mini

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Originally posted by Gruma
Could someone solve this question as I think the answer in the textbook is wrong.

30ml of 0.374 mol L -1 sulphuric acid is diluted so the concentration becomes 0.110 mol L-1. The volume of water that must be added is:
Method 1: Use the ratio of concentrations to find the ratio of volumes. This gives a 3.4-fold dilution. So 102mL is the new volume and 72 extra mL needed

Method 2: Equate the moles of acid. This gives 102mL as the new volume, so 72mL were added.
 

Gruma

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Chemistry q

Thanks
I just made a stupid mnistake at the end I did use the same method c1 v1 =c 2 v2
 

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