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Principles of Instrumental Analysis, 6th ed. Chapter 24
 
 5
 1 mol Tl 4 mol e 1 F 96485 C = 0.300 g 
204.37 g 2 mol Tl mol e F
Q
−
−× × × × =283.27 C 
 t = 283.27 C/0.905 C/s = 313.00 s or 5.22 min 
 (c) Proceeding as in part (a), with the only difference being 1 mol e–/mol Tl 
 Q = 141.63 C 
 t = 156.5 s or 2.61 min 
24-8. Sample 1 at – 1.0V 42 mol CCl1 mol e11.63 C 
96485 C 2 mol e
−
−× × = 1.25 × 10–4mol CCl4
 at –1.80 V 32 mol CHCl1 mol e68.60 C 
96485 C 6 mol e
−
−× × = 2.370 × 10–4 mol CHCl3
 original no. mol CHCl3 = 2.370 × 10–4 – 1.205 × 10–4 = 1.165 × 10–4
 
4
4 4 41.205 10 mol CCl 153.82 g CCl / mol CCl 100%
0.750 g sample
−× ×
× = 2.47% CCl4
 
4
3 3 31.165 10 mol CHCl 119.37 g CHCl / mol CHCl 100%
0.750 g sample
−× ×
× = 1.85% CHCl3 
In a similar manner, we can calculate the other percentages as shown in the spreadsheet.