CHEMISTRY, M2 EQ-Bank 14

The concentration of a solution of hydrochloric acid \(\ce{HCl}\) is 1.80% (w/v). What is the molar concentration produced by diluting 20.0 mL of this solution to a total volume of 200.0 mL with deionised water? (3 marks)

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\(0.049\ \text{mol/L}\)

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→ \(1.8\)% (w/v) means that there is \(1.8\ \text{g}\) of solute in \(100\ \text{mL}\).

→ In \(20\ \text{mL}\), there is \(1.8\ \text{g} \times 0.2 = 0.36\ \text{g}\) of \(\ce{HCl}\).

\(n\ce{(HCl)} = \dfrac{0.36}{1.008 + 35.45} = 0.00987\ \text{mol}\)

→ The final concentration after the dilution will be:

\(c = \dfrac{n}{V} = \dfrac{0.00987}{0.2} = 0.049\ \text{mol/L}\)