Answer to Question #88007 in Inorganic Chemistry for Rafsan

Question #88007
What weight of HCl is present in 155 ml of a 0.540 M solution ?
1
Expert's answer
2019-04-15T07:54:49-0400

Vs(HCl) = 155 mL = 0.155 L.

n(HCl) = Vs(HCl) × C(HCl) = 0.155 L × 0.540 M = 0.0837 mol.

m(HCl) = n(HCl) × M(HCl) = 0.0837 mol × 36.5 g/mol = 3.05505 g ≈ 3.055 g.


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