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# BMAT section 2 help! watch

1. I am stuck in one of the section 2 question. Can anyone help me out?
They've got an explanation with it but I just don't get where is the 1.005 comes from.

Q) A student is using a volumetric pipette to titrate a solution of H2SO4. The pipette volume is reported as 1.00ml, and the titration requires 12 pipette volumes of KOH solution to neutralise 1 pipette volume of H2SO4 solution.
If the KOH solution concentration is reported as 0.100 mol/dm3, which of the following is an expression for the maximum possible concentration (in mol/dm3) of the solution of H2SO4 that can be calculated from this data?

Ans : 6 x 1.005 x 0.1005
0.955

Explnation: The question involves a neutralisation reaction. The KOH solution concentration is 0.100 moldm–3 and the volume of this solution is 12 x 1.00 mL. The H2SO4 solution volume is 1 x 1.00 mL. The question asks the maximum possible concentration of the H2SO4 solution. The maximum concentration of the H2SO4 solution would be calculated from the minimum possible volume of H2SO4 solution and the maximum possible volume of KOH solution of maximum possible concentration.
2KOH + H2SO4 → 2H2O + K2SO4
So there are two equivalents of OH– for every equivalent of H2SO4.
Now work out the numbers of moles involved in the neutralisation:
Volume of KOH = 12 x 1.005 mL, concentration of KOH = 0.1005 moldm–3

where does the 0.1005 comes from? I thought KOH concentration is 0.100 and volume is 12/1000 = 0.012dm3.

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