hydrated salts maths question
Please help me with this question type I can’t understand the portions bit of it and the molar mass things.
A 200.0 cm3 standard solution of a hydrate contains 8.50 g of MgSO4⋅xH2O. 25.0 cm3 portions of this solution are titrated with a 0.250 mol/dm potassium permanganate solution. The mean titre is 30.0 cm3. Calculate the value of x, the number of water molecules of crystallization.
A 200.0 cm3 standard solution of a hydrate contains 8.50 g of MgSO4⋅xH2O. 25.0 cm3 portions of this solution are titrated with a 0.250 mol/dm potassium permanganate solution. The mean titre is 30.0 cm3. Calculate the value of x, the number of water molecules of crystallization.
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by Irijxx111Please help me with this question type I can’t understand the portions bit of it and the molar mass things.
A 200.0 cm3 standard solution of a hydrate contains 8.50 g of MgSO4⋅xH2O. 25.0 cm3 portions of this solution are titrated with a 0.250 mol/dm potassium permanganate solution. The mean titre is 30.0 cm3. Calculate the value of x, the number of water molecules of crystallization.
A 200.0 cm3 standard solution of a hydrate contains 8.50 g of MgSO4⋅xH2O. 25.0 cm3 portions of this solution are titrated with a 0.250 mol/dm potassium permanganate solution. The mean titre is 30.0 cm3. Calculate the value of x, the number of water molecules of crystallization.
Suppose you use only 25 cm^3 of the 200 cm^3 of solution. 25 cm^3 is 25/200 or 1/8 of the solution and so you have only reacted 1/8 of the stuff dissolved in it (e.g 8.50/8 g = 1.0625 g of MgSO4.xH2O) in the titration.
I don’t think this particular example can be worked through as MgSO4.xH2O cannot be oxidised and the presumable products of this reaction are soluble (so you don’t get a precipitate), though. Had it been MgSO3.xH2O with acidified KMnO4, it would be doable as the sulphate(IV) ion is oxidised as follows:
5SO3^2- + 2MnO4^- + 6H^+ —> 5SO4^2- + 2Mn^2+ + 3H2O
(edited 11 months ago)
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