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Calculate the mass of glucose needed to make 250 mL of a 0.4 M solution, and the molality of a solution made with 45 g of glucose in 500 g of water. Finally, what is the infant dose of a medication gi

Chemistry

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5 parts
a

18.016 g\text{18.016 g}

b

0.5 mol/kg\text{0.5 mol/kg}

c

0.5332 mL\text{0.5332 mL}

d

0.4 ppm\text{0.4 ppm}

e

142.86 g of 15% zinc oxide ointment\text{142.86 g of 15\% zinc oxide ointment}

107.14 g of 8% zinc oxide ointment\text{107.14 g of 8\% zinc oxide ointment}

Step-by-step solution

12 steps

Part a· 2 steps

  1. Calculate the moles of glucose needed.

    Molarity (MM) is moles of solute per liter of solution.

    Moles=M×V\text{Moles} = M \times V Moles=0.4 molL×0.250 L\text{Moles} = 0.4 \, \frac{\text{mol}}{\text{L}} \times 0.250 \, \text{L} Moles=0.1 mol\text{Moles} = 0.1 \, \text{mol}

  2. Calculate the mass of glucose in grams.

    Mass=Moles×MW\text{Mass} = \text{Moles} \times \text{MW} Mass=0.1 mol×180.16 gmol\text{Mass} = 0.1 \, \text{mol} \times 180.16 \, \frac{\text{g}}{\text{mol}} Mass=18.016 g\text{Mass} = 18.016 \, \text{g}

    The mass of glucose needed is 18.016 g\boxed{\text{18.016 g}}.

Part b

0.5 mol/kg\text{0.5 mol/kg}

  1. Calculate the moles of glucose (solute).

    Moles of glucose=Mass of glucoseMW of glucose\text{Moles of glucose} = \frac{\text{Mass of glucose}}{\text{MW of glucose}} Moles of glucose=45 g180 gmol\text{Moles of glucose} = \frac{45 \, \text{g}}{180 \, \frac{\text{g}}{\text{mol}}} Moles of glucose=0.25 mol\text{Moles of glucose} = 0.25 \, \text{mol}

  2. Convert the mass of water (solvent) to kilograms.

    Mass of water=500 g=0.500 kg\text{Mass of water} = 500 \, \text{g} = 0.500 \, \text{kg}

  3. Calculate the molality (mm).

    Molality is moles of solute per kilogram of solvent.

    m=Moles of soluteMass of solvent (kg)m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} m=0.25 mol0.500 kgm = \frac{0.25 \, \text{mol}}{0.500 \, \text{kg}} m=0.5 molkgm = 0.5 \, \frac{\text{mol}}{\text{kg}}

    The molality of the solution is 0.5 mol/kg\boxed{\text{0.5 mol/kg}}.

Part c

0.5332 mL\text{0.5332 mL}

  1. Calculate the infant dose using Fried's Rule.

    Fried's Rule: Infant Dose = Age in months150×Adult Dose\frac{\text{Age in months}}{150} \times \text{Adult Dose}

    Infant Dose=10 months150×200 mg\text{Infant Dose} = \frac{10 \, \text{months}}{150} \times 200 \, \text{mg} Infant Dose=115×200 mg\text{Infant Dose} = \frac{1}{15} \times 200 \, \text{mg} Infant Dose=13.33 mg\text{Infant Dose} = 13.33 \, \text{mg}

  2. Calculate the volume of suspension to administer.

    The suspension concentration is 125 mg/5 mL.

    Volume=Infant DoseConcentration\text{Volume} = \frac{\text{Infant Dose}}{\text{Concentration}} Volume=13.33 mg125 mg5 mL\text{Volume} = \frac{13.33 \, \text{mg}}{125 \, \frac{\text{mg}}{5 \, \text{mL}}} Volume=13.33 mg125 mg×5 mL\text{Volume} = \frac{13.33 \, \text{mg}}{125 \, \text{mg}} \times 5 \, \text{mL} Volume=0.10664×5 mL\text{Volume} = 0.10664 \times 5 \, \text{mL} Volume=0.5332 mL\text{Volume} = 0.5332 \, \text{mL}

    The dose to be administered is 0.5332 mL\boxed{\text{0.5332 mL}}.

Part d

0.4 ppm\text{0.4 ppm}

  1. Convert the volume of solution to liters.

    Volume of solution=2,000 mL=2 L\text{Volume of solution} = 2,000 \, \text{mL} = 2 \, \text{L}

  2. Calculate the concentration in parts per million (ppm).

    For dilute aqueous solutions, ppm can be calculated as mg of solute per liter of solution.

    ppm=Mass of lead (mg)Volume of solution (L)\text{ppm} = \frac{\text{Mass of lead (mg)}}{\text{Volume of solution (L)}} ppm=0.8 mg2 L\text{ppm} = \frac{0.8 \, \text{mg}}{2 \, \text{L}} ppm=0.4 ppm\text{ppm} = 0.4 \, \text{ppm}

    The concentration in ppm is 0.4 ppm\boxed{\text{0.4 ppm}}.

Part e

142.86 g of 15% zinc oxide ointment\text{142.86 g of 15\% zinc oxide ointment}

  1. Use alligation to find the ratio of the two ointments.

    We want to prepare 12% zinc oxide ointment using 8% and 15% ointments.
    Draw the alligation grid:

    15% (Higher)       |       4 parts (12 - 8) of 15% ointment
                   |
        12% (Target)
                   |
 8% (Lower)        |       3 parts (15 - 12) of 8% ointment

    The ratio of 15% ointment to 8% ointment is 4:3.
    Total parts = 4+3=74 + 3 = 7 parts.

  2. Calculate the amount of 15% zinc oxide ointment needed.

    \text{Amount of 15% ointment} = \frac{4 \, \text{parts}}{7 \, \text{parts}} \times 250 \, \text{g} \text{Amount of 15% ointment} = \frac{4}{7} \times 250 \, \text{g} \text{Amount of 15% ointment} \approx 142.86 \, \text{g}

  3. Calculate the amount of 8% zinc oxide ointment needed.

    \text{Amount of 8% ointment} = \frac{3 \, \text{parts}}{7 \, \text{parts}} \times 250 \, \text{g} \text{Amount of 8% ointment} = \frac{3}{7} \times 250 \, \text{g} \text{Amount of 8% ointment} \approx 107.14 \, \text{g}

    To prepare 250g of 12% zinc oxide ointment, you need \boxed{\text{142.86 g of 15% zinc oxide ointment}} and \boxed{\text{107.14 g of 8% zinc oxide ointment}}.

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