Question

what weight of HCl is present in 155ml of 0.540 M solution​

Answer 1

Answer:

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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Answer 2

Explanation:

3.055 gram hope it'll help you make it brainliest answer please