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A student is given a solution A of hydrochloric acid (HCl) of unknown concentration and a solution B of an alkali XOH of unknown concentration. They are asked to determine the concentrations of these

Chemistry

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2 parts
a(i)

0.104 mol dm−3\text{0.104 mol dm}^{-3}

a(iv)

0.076 g\text{0.076 g}

Step-by-step solution

4 steps

To solve this problem, we need to use the given information and make an assumption for the atomic mass of chlorine, as it is not provided.

Assumption: The atomic mass of Chlorine (Cl\text{Cl}) is 35.5.

Part a(i)· 2 steps

Concentration of A (HCl) in mol dm−3\text{mol dm}^{-3}

  1. Calculate the molar mass of HCl.

    The atomic mass of H is 1 g mol−11 \, \text{g mol}^{-1}.
    The atomic mass of Cl is assumed to be 35.5 g mol−135.5 \, \text{g mol}^{-1}.

    Molar mass of HCl=Atomic mass of H+Atomic mass of Cl\text{Molar mass of HCl} = \text{Atomic mass of H} + \text{Atomic mass of Cl} Molar mass of HCl=1 g mol−1+35.5 g mol−1=36.5 g mol−1\text{Molar mass of HCl} = 1 \, \text{g mol}^{-1} + 35.5 \, \text{g mol}^{-1} = 36.5 \, \text{g mol}^{-1}
  2. Calculate the molar concentration of HCl (Solution A).

    The mass concentration of HCl is given as 3.80 g dm−33.80 \, \text{g dm}^{-3}.

    Molar concentration=Mass concentrationMolar mass\text{Molar concentration} = \frac{\text{Mass concentration}}{\text{Molar mass}} Concentration of A (HCl)=3.80 g dm−336.5 g mol−1\text{Concentration of A (HCl)} = \frac{3.80 \, \text{g dm}^{-3}}{36.5 \, \text{g mol}^{-1}} Concentration of A (HCl)≈0.1041 mol dm−3\text{Concentration of A (HCl)} \approx 0.1041 \, \text{mol dm}^{-3}

    Rounding to three significant figures:

    0.104 mol dm−3\boxed{\text{0.104 mol dm}^{-3}}

Part a(ii)

Concentration of B in mol dm−3\text{mol dm}^{-3}

Part a(iii)

Calculate the molar mass of XOH

These parts require experimental results from the titration, specifically the average volume of solution A used (the titre volume). Since no burette readings or average titre volume are provided, these calculations cannot be completed.

Part a(iv)

0.076 g\text{0.076 g}

Mass of HCl in 20 cm320 \, \text{cm}^3 of the solution.

  1. Convert the given volume from cm3\text{cm}^3 to dm3\text{dm}^3.

    20 cm3=20×10−3 dm3=0.020 dm320 \, \text{cm}^3 = 20 \times 10^{-3} \, \text{dm}^3 = 0.020 \, \text{dm}^3
  2. Calculate the mass of HCl using its mass concentration.

    The mass concentration of HCl is given as 3.80 g dm−33.80 \, \text{g dm}^{-3}.

    Mass of HCl=Mass concentration×Volume\text{Mass of HCl} = \text{Mass concentration} \times \text{Volume} Mass of HCl=3.80 g dm−3×0.020 dm3\text{Mass of HCl} = 3.80 \, \text{g dm}^{-3} \times 0.020 \, \text{dm}^3 Mass of HCl=0.076 g\text{Mass of HCl} = 0.076 \, \text{g} 0.076 g\boxed{\text{0.076 g}}
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