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Chapter 13: Problem 109

The concentration of \(\mathrm{N}_{2}\) in the ocean at \(25^{\circ} \mathrm{C}\) is \(445 \mu \mathrm{M} .\) The Henry's law constant for \(\mathrm{N}_{2}\) is \(0.61 \times 10^{-3} \mathrm{mol} \mathrm{L}^{-1} \mathrm{atm}^{-1} .\) Calculate the mass of \(\mathrm{N}_{2}\) in a liter of ocean water. Calculate the partial pressure of \(\mathrm{N}_{2}\) in the atmosphere.

Short Answer

Expert verified
There are approximately 0.01246 grams of \(\mathrm{N}_{2}\) in a liter of ocean water and the partial pressure of \(\mathrm{N}_{2}\) is about 0.000271 atm.

Step by step solution

01

Calculate the Molar Mass of N2

First, based on the given concentration of \(\mathrm{N}_{2}\), we need to calculate how many moles of \(\mathrm{N}_{2}\) are in one liter of sea water. The given concentration is \(445 \mu \mathrm{M}\), which equals \(445 \times 10^{-6} \mathrm{mol/L}\). This means there are \(445 \times 10^{-6} \mathrm{mol}\) of \(\mathrm{N}_{2}\) per liter of ocean water.
02

Determine the Mass of N2

Once we know the concentration in moles, we can find out the mass of \(\mathrm{N}_{2}\) per liter. Given the molar mass of \(\mathrm{N}_{2}\) is 28 grams per mole, we can multiply this value by the number of moles from the first step: \(445 \times 10^{-6} \mathrm{moles} \times 28 \mathrm{g/mole} = 0.01246 \mathrm{g}\). So, there are approximately 0.01246 g of \(\mathrm{N}_{2}\) in a liter of ocean water.
03

Calculate the Partial Pressure of N2

To find the partial pressure of \(\mathrm{N}_{2}\), we can use the equation of Henry's law: \(P = C \times K\), where \(P\) is the partial pressure, \(C\) is the concentration, and \(K\) is Henry's law constant. We substitute the given values into the equation: \(P = 445 \times 10^{-6} \mathrm{mol/L} \times 0.61 \times 10^{-3} \mathrm{ atm/L}\). Upon solving, we find that the partial pressure of \(\mathrm{N}_{2}\) is approximately \(0.000271 \mathrm{atm}\).

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Most popular questions from this chapter

The aqueous solubility at \(20^{\circ} \mathrm{C}\) of \(\mathrm{Ar}\) at \(1.00 \mathrm{atm}\) is equivalent to \(33.7 \mathrm{mL} \mathrm{Ar}(\mathrm{g}),\) measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1.00 atm and \(20^{\circ} \mathrm{C}\) ? Air contains \(0.934 \%\) Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.

At \(25^{\circ} \mathrm{C}\) a \(0.50 \mathrm{g}\) sample of polyisobutylene (a polymer used in synthetic rubber) in \(100.0 \mathrm{mL}\) of benzene solution has an osmotic pressure that supports a \(5.1 \mathrm{mm}\) column of solution \((d=0.88 \mathrm{g} / \mathrm{mL}) .\) What is the molar mass of the polyisobutylene? (For \(\mathrm{Hg}\), \(d=13.6 \mathrm{g} / \mathrm{mL} .)\)

Styrene, used in the manufacture of polystyrene plastics, is made by the extraction of hydrogen atoms from ethylbenzene. The product obtained contains about \(38 \%\) styrene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2}\right)\) and \(62 \%\) ethylbenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3}\right),\) by mass. The mixture is separated by fractional distillation at \(90^{\circ} \mathrm{C} .\) Determine the composition of the vapor in equilibrium with this \(38 \%-62 \%\) mixture at \(90^{\circ} \mathrm{C}\). The vapor pressure of ethylbenzene is \(182 \mathrm{mmHg}\) and that of styrene is \(134 \mathrm{mmHg}\).

Calculate the vapor pressure at \(25^{\circ} \mathrm{C}\) of a solution containing \(165 \mathrm{g}\) of the nonvolatile solute, glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in \(685 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8 \mathrm{mmHg}\).

Calculate the vapor pressure at \(20^{\circ} \mathrm{C}\) of a saturated solution of the nonvolatile solute, urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) in methanol, \(\mathrm{CH}_{3} \mathrm{OH} .\) The solubility is \(17 \mathrm{g}\) urea/100 \(\mathrm{mL}\) methanol. The density of methanol is \(0.792 \mathrm{g} / \mathrm{mL}\), and its vapor pressure at \(20^{\circ} \mathrm{C}\) is \(95.7 \mathrm{mmHg}\).
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