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Answer to Question #175791 in General Chemistry for Kiana
Question #175791
A 4.0520g sample of HCl, sp. Gr
11.18, required 44.15ml of 0.9035M
of Sodium Hydroxide in titration,
compute for percent purity.
11.18, required 44.15ml of 0.9035M
of Sodium Hydroxide in titration,
compute for percent purity.
Expert's answer
44.15 x 0.9035/1000 = 0.04.
0.04 x 36.5 = 1.46 g of HCl.
1.46 x 100/4.052 = 36,0315893 %
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