Question

309 Week One Data and Calculations Standardization of NaOH Solution Sample 1 Sample 2 Sample 3 Mass of KHP 10.5749 10.8939 1.26g. Initial Volume NaOH, V. (mL) Read to nearest 0.01 ml. No.25mL 8.86 m L 10.45 mL Final Volume NaOH, V:(mL) 0.31 ml 25.73 m2 Read to nearest 0.01 mL 24.47 ml Total Volume NaOH used mL) 110.62 mL 16.92 mL 24.03 mL Moles KHP 2.81 × 10-3 4.37x10-3 6.16 x 10 3 Moles NaOH 12.81x 10-3 4.37 × 103 6.16*103 Molarity NaOH 10. 264M 0.250 0.256M.) Average Molarity 0.259. Standard Deviation on M 4.183 x 10-3

Answer 1

Chemical Reavian of Kip and Noor хои + Кися ИО, kwa Cen Ou + H, O Kup and Naon React in 1:1 so mole af kup = mole of Naoh 0.574 g = 2.81X103 = mole of Nood mole af kHP = 204.22 g/mol blittre 10.62 ml = 0.010 62 Naon volume = Molarity = No of moles Volume in (L) 0.264 m 2.81 x 103 0.01062 -= Sampl-2 Mole of Kup = 0.893g = 4.37x10s = male of Nach 204.22 g - mod (12=1000 me) 0.01692L Naon volume = 16.92 me = Naon Mularity = 4.37810S -=012582M 0.01692 sample-3 126 = 6.168103 = Mole of Naast mole of KHP = - 204.22 = 6.168103 am Volume af Naon = 24.03 ml = 0.02403L Molality - 616810-3 0.02403 = 0.256 m 0.259 m Qu264 + 0.2582 +0.256 Average molarity of Naol = 0 3 - Yes your calculation is connect.