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What mass of zinc sulfate crystals (ZnSO4·7H2O) is needed to provide 15 mg of zinc?

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2 parts
i

66.3 mg\text{66.3 mg}

ii

0.0462 mol/dm3\text{0.0462 mol/dm}^3

Step-by-step solution

8 steps

Part i· 5 steps

What mass of zinc Sulphate crystals would need to be taken to obtain this intake?

  1. Convert the daily zinc intake from milligrams to grams.

    Daily zinc intake=15 mg=15×10−3 g\text{Daily zinc intake} = 15 \text{ mg} = 15 \times 10^{-3} \text{ g}

  2. Calculate the moles of zinc (Zn\text{Zn}) required.

    The atomic mass of zinc (Zn\text{Zn}) is 65 g/mol65 \text{ g/mol}.

    Moles of Zn=Mass of ZnMolar mass of Zn\text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} Moles of Zn=15×10−3 g65 g/mol≈0.00023077 mol\text{Moles of Zn} = \frac{15 \times 10^{-3} \text{ g}}{65 \text{ g/mol}} \approx 0.00023077 \text{ mol}

  3. Determine the moles of hydrated zinc sulfate (ZnSO4⋅7H2O\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}) needed.

    From the formula ZnSO4⋅7H2O\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}, 1 mole of the hydrated salt contains 1 mole of zinc.

    Moles of ZnSO4⋅7H2O=Moles of Zn≈0.00023077 mol\text{Moles of ZnSO}_4 \cdot 7\text{H}_2\text{O} = \text{Moles of Zn} \approx 0.00023077 \text{ mol}

  4. Calculate the molar mass of hydrated zinc sulfate (ZnSO4⋅7H2O\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}).

    Atomic masses: Zn=65\text{Zn} = 65, S=32\text{S} = 32, O=16\text{O} = 16, H=1\text{H} = 1.

    Molar mass of ZnSO4=65+32+(4×16)=161 g/mol\text{Molar mass of ZnSO}_4 = 65 + 32 + (4 \times 16) = 161 \text{ g/mol} Molar mass of 7H2O=7×((2×1)+16)=7×18=126 g/mol\text{Molar mass of } 7\text{H}_2\text{O} = 7 \times ((2 \times 1) + 16) = 7 \times 18 = 126 \text{ g/mol} Molar mass of ZnSO4⋅7H2O=161+126=287 g/mol\text{Molar mass of ZnSO}_4 \cdot 7\text{H}_2\text{O} = 161 + 126 = 287 \text{ g/mol}

  5. Calculate the mass of hydrated zinc sulfate crystals needed.

    Mass=Moles×Molar mass\text{Mass} = \text{Moles} \times \text{Molar mass} Mass=0.00023077 mol×287 g/mol≈0.06630 g\text{Mass} = 0.00023077 \text{ mol} \times 287 \text{ g/mol} \approx 0.06630 \text{ g}

    Converting to milligrams:

    Mass=0.06630 g×1000 mg/g=66.3 mg\text{Mass} = 0.06630 \text{ g} \times 1000 \text{ mg/g} = 66.3 \text{ mg}

    The mass of zinc sulfate crystals needed is 66.3 mg\boxed{\text{66.3 mg}}.

Part ii

0.0462 mol/dm3\text{0.0462 mol/dm}^3

Calculate the concentration of this solution in mol/dm3^3 of the hydrated salt.

  1. Identify the moles of hydrated zinc sulfate.

    From part (i), the moles of ZnSO4⋅7H2O\text{ZnSO}_4 \cdot 7\text{H}_2\text{O} are approximately 0.00023077 mol0.00023077 \text{ mol}.

  2. Convert the volume of the dose from millilitres to litres (dm3^3).

    Volume=5 ml=5 cm3=0.005 L=0.005 dm3\text{Volume} = 5 \text{ ml} = 5 \text{ cm}^3 = 0.005 \text{ L} = 0.005 \text{ dm}^3

  3. Calculate the concentration of the solution.

    Concentration=Moles of hydrated saltVolume of solution (in dm3)\text{Concentration} = \frac{\text{Moles of hydrated salt}}{\text{Volume of solution (in dm}^3)} Concentration=0.00023077 mol0.005 dm3≈0.046154 mol/dm3\text{Concentration} = \frac{0.00023077 \text{ mol}}{0.005 \text{ dm}^3} \approx 0.046154 \text{ mol/dm}^3

    Rounding to three significant figures:
    The concentration of the solution is 0.0462 mol/dm3\boxed{\text{0.0462 mol/dm}^3}.

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