Answer to Question #88007 in Inorganic Chemistry for Rafsan

Inorganic Chemistry

Question #88007

What weight of HCl is present in 155 ml of a 0.540 M solution ?

Expert's answer

V_s(HCl) = 155 mL = 0.155 L.

n(HCl) = V_s(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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