Chemistry Question (1 Viewer)

[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:

 

[wogboy]

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.

 

[kini mini]

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]

Chemistry q

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