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NJC_H2_CHEM_P4_Solutions.pdf - Suggested solution for 2017...

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Suggested solution for 2017 SH 2 Prelim Practical 1 (a) (ii) Table 1: Weighings of FA1 Mass of weighing bottle and FA1 / g 3.696 Mass of weighing bottle and residual FA1 / g 3.497 Mass of FA1 transferred / g 0.199 Table 2: Titration of FA3 with FA4 using screened methyl orange as indicator 1 2 Final burette reading / cm 3 30.00 36.90 Initial burette reading / cm 3 9.00 16.00 Volume of FA4 used / cm 3 21.00 20.90 (iii) average volume = 21.00 + 20.90 2 = 20.95 cm 3 (b) (i) amount of HNO 3 in 25.0 cm 3 = amount of NaOH = 20.95 / 1000 x (9.10/40.0) = 4.77 x 10 3 amount of HNO 3 in 250.0 cm 3 = 4.77 x 10 2 (ii) amount of HNO 3 in 25.0 cm 3 = 25/1000 x 2 = 0.05 mol (iii) amount of HNO 3 = 0.05 ‒ 4.77 x 10 2 = 2.34 x 10 3 (iv) amount of MgCO 3 .xH 2 O = 2.34 x 10 3 / 2 = 1.17 x 10 3 M r = 0.208 / 1.15 x 10 3 = 170.2 x = [180.7 (24.3 + 12.0 + 16.0 x 3)] / 18 = 4.77 ≈ 5 (c) The acid will not be diluted and close to 2.00 mol dm 3 , the titre value required to neutralise 25 cm 3 of FA 3 will be about 250 cm 3 / more than 10 times greater . Since this exceeds the capacity of burette , it is unsuitable.
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