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Chapter 13: Problem 23

What is the molarity of \(\mathrm{CO}_{2}\) in a liter of ocean water at \(25^{\circ} \mathrm{C}\) that contains approximately \(280 \mathrm{ppm}\) of \(\mathrm{CO}_{2} ?\) The density of ocean water is \(1027 \mathrm{kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The molarity of CO2 in ocean water at 25 °C containing approximately 280 ppm of CO2 is 0.00636 M.

Step by step solution

01

Understand the given information

The ocean water contains approximately 280 parts per million (ppm) of CO2. This means for every 1 million parts of solution, there are 280 parts of CO2. The density of ocean water is given as 1027 kg/m^3.
02

Convert ppm unit to grams

280 ppm of CO2 is equivalent to 280 mg of CO2 in 1kg (or 1 liter) of water. To have the amount of CO2 in grams, divide by 1000: \(280 \, \text{ppm} = 280 \, \text{mg/kg} = 280/1000 \, \text{g} = 0.28 \, \text{g}\)
03

Calculate moles of CO2

Use the molar mass of CO2 to convert grams to moles. The molar mass of CO2 is approximately 44.01 g/mol. Divide the mass by molar mass: \(0.28 \, \text{g} / 44.01 \, \text{g/mol} = 0.00636 \, \text{moles}\)
04

Calculate molarity

Finally, calculate the molarity of CO2 by using the definition of molarity, which is the moles of solute per liter of solution. Hence divide the moles by the volume in liters. As 1 kg of water is approximately equal to 1 liter, divide the moles by 1 liter to obtain the molarity: \(0.00636 \, \text{moles/L} = 0.00636 \, \text{M}\)

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Most popular questions from this chapter

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