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.
1
Expert's answer
2021-03-29T05:19:17-0400

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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