Answer to Question #176040 in General Chemistry for Alend

Question #176040

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.

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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