What weight of HCl is present in 155 ml of a 0.54 M solution?

Question

What weight of HCl is present in 155 ml of a 0.54 M solution? (A) 3.06 g (B) 6.12 g (C) 1.53 g (D) 0.30

Tardigrade Admin · Accepted Answer

HCl; molecular weight = 36.4 g/mol M = (g/mw)/ L M ⋅ L= g/m w mw × m × L = g g = (36.4 g/mol)(0.54 mol/L) (155 mL) (1L/1000 mL) g = 3.04 g ≃ 3.06 g

Q. What weight of $H Cl$ is present in $155 m l$ of a $0.54 M$ solution?