From the following data, find the % (w/w) cream of tartar (KHC4H406. MM=188): Wt of sample = 1.4160 g N2OH titrant used = 20.87 mL HCl used for back titration = 1.27 mL 1.00 mL HCI = 1.12 mL N2OH 1.00 mL HCI = 0.01930 g CaCO3 O 95.59% (w/w) O 89.08% (w/w) O 47.75% (w/w) O 87.12% (w/w)

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QUESTION 29
From the following data, find the % (w/w) cream of tartar (KHC4H406, MM=188):
Wt of sample = 1.4160 g
NaOH titrant used = 20.87 mL
HCI used for back titration = 1.27 mL
1.00 mL HCI = 1.12 mL NaOH
1.00 mL HCI = 0.01930 g Caco3
95.59% (w/w)
89.08% (w/w)
O 47.75% (w/w)
O 87.12% (w/w)
Transcribed Image Text:QUESTION 29 From the following data, find the % (w/w) cream of tartar (KHC4H406, MM=188): Wt of sample = 1.4160 g NaOH titrant used = 20.87 mL HCI used for back titration = 1.27 mL 1.00 mL HCI = 1.12 mL NaOH 1.00 mL HCI = 0.01930 g Caco3 95.59% (w/w) 89.08% (w/w) O 47.75% (w/w) O 87.12% (w/w)
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