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Mrac [35]
2 years ago
6

Wastewater from a cement factory contains 0.410 g of Ca2+ ion and 0.0330 g of Mg2+ ion per 100.0 L of solution. The solution den

sity is 1.001 g/mL. Calculate the Ca2+ and Mg2+ concentrations in ppm (by mass).
Chemistry
1 answer:
Delvig [45]2 years ago
3 0

Answer:

Ca^{2+}= 4.0959 ppm

Mg^{2+}= 0.3296 ppm

Explanation:

ppm means milligram of solute per kilogram of solution.

ppm= \frac{mg of solute}{Kg of solution}

Solutes: Ca^{2+} y Mg^{2+}

Mass Ca^{2+} = 0.410g

Mass Mg^{2+} = 0.0330g

Solution: 100.0 L, whose density is 1.001 g/mL

First, let us to calculate mass of solution:

100.0 L * \frac{1.001 g }{1 mL } * \frac{1000 mL}{1 L} * \frac{1 kg}{1000 g} == 100.1 kg

Now, we have to calculate ppm Ca^{2+} y Mg^{2+}

ppm Ca^{2+}= \frac{0.410 g * (1000 mg/1g)}{100.1 Kg} =4.0959 mg/Kg

ppm Mg^{2+}= \frac{0.0330 g * (1000 mg/1g)}{100.1 Kg} =0.3296 mg/Kg

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