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Seawater has a concentration of 0.60 M NaCl. The acceptable amount of NaCl in drinking water is 9.0 g/L. How much pure water

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Answer #1

Moles of NaCl in 1 L of 0.6 M water = Volume * molarity = 1 * 0.6 = 0.6 moles

Mass of Sodium = Moles * molar mass = 0.6 * 58.5 = 35.1

Present concentration = 35.1 g/L of NaCl

Assuming that 2.5 L of the original water sample is diluted to a final volume V with concentration 9 g/L, we use the relation:

C * V = constant, where c is concentration

35.1 * 2.5 = 9 * V

Solving, V = 9.75 L

Water added = 9.75 - 2.5 = 7.25 L

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