I want part and 3 please Part l: Unit Conversions 1. The EPA has set a Maximum C

Question

I want part and 3 please

Part l: Unit Conversions 1. The EPA has set a Maximum Contaminant Level (MCL) of 10 ppm for nitrate in drinking water. What is that MCL in moles per liter of nitrate? have a water sample that contains 1.8% sodium and 12,402 ppm chloride. Which is more abundant in this water sample, the sodium or the chloride? 3. At 25°C, the solubility limit of calcium carbonate is 0.013 g L (you could dissolve up to that much before calcium carbonate will start to precipitate from the solution). What is the maximum percentage of calcium carbonate that we could dissolve in water at 25 C?

Explanation / Answer

1) The MCL of nitrate, NO3- in drinking water is 10 ppm.

Since, 1 ppm = 1 mg/L, we have the MCL of NO3- = 10 ppm = 10 mg/L.

Molar mass of NO3- = (1*14.0067 + 3*15.9994) g/mol = 62.0049 g/mol.

Therefore, we have MCL of NO3- = (10 mg/L)*(1 g/1000 mg)*(1 mole/62.0049 g) = 1.613*10-4 mol/L (ans).

2) We have 1.8% sodium and 12,402 ppm chloride in a water sample.

We know that 1.8% sodium = 1.8 g sodium/100 mL water sample. Therefore, the mass of sodium present in 1 L =1000 mL of the water sample = (1.8 g/100 mL)*(1000 mL) = 18 g.

We know that 1 ppm = 1 mg/L; therefore, the chloride concentration is 12,402 ppm = 12,402 mg/L = (12,402 mg/L)*(1 g/1000 mg) = 12.402 g/L.

Therefore, the sodium and chloride concentrations are 18 g/L and 12.402 g/L. Consequently, the water sample has a higher sodium concentration (ans).

3) We have the solubility of calcium carbonate in water as 0.013 g/L at 25°C. Therefore, we have 1 L = 1000 mL water dissolve 0.013 g calcium carbonate.

100 mL water will dissolve (0.013 g/1000 mL)*(100 mL) = 0.0013 g calcium carbonate.

We define the percent solubility of a compound in water as the grams of the compound in 100 mL water; hence, the percentage solubility of calcium carbonate in water at 25°C is 0.0013% (ans).

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