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# Titration Calculation! watch

1. Hello TSR,

I was wondering if someone could tell me where i have gone wrong here please:-

Caclulate the molar mass of the acid H2(X)

A student dissolved 1.571g of an acid, H2(X), in water and made the solution up to 250cm^3.

She titrated 25.0cm^3 of this solution against 0.125mol dm^-3 sodium hydroxide, NaOH. 21.30cm^3 of NaOH were needed to reach the end point.

The equation for this reaction is:-

2NaOH + H2(X) - - - - Na2(X) + 2H2(O)

My Calculation:-

n = 0.125 * (21.30/1000) = 2.6625*10^-3 mol.

(2.6625*10^-3)/2 = 1.33125*10^-3 mol. H2(X)

M = (1.571g)/(1.33125*10^-3) = 1180.093897 g mol -1.

The answer is 118 g mol -1, but i got 1180.093897... could anyone please tell me where i have gone wrong??

Much Appreciated!! Thanks!
2. you have to divide the answer by 10 because remember - she made the solution up to 250cm^3 then used 25cm^3 of it.
3. Oh right, thank you for that!!

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