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Chapter 13: Problem 47

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.

Short Answer

Expert verified
The molarity of Ar in water that is saturated with air at 1.00 atm and \( 20^{\circ} \mathrm{C} \) is 0.000417 M.

Step by step solution

01

Find volume of Ar in air

Air contains 0.934% Ar by volume. This means, in 1000 mL (or 1 L) of air, there will be 0.934% of 1000 mL = 0.00934 * 1000 mL = 9.34 mL of Ar.
02

Convert volume of Ar to moles

At STP, 1 mole of any gas occupies 22.4 L or 22400 mL. So, 9.34 mL of Ar is equivalent to \( \frac{9.34}{22400} \) moles = 0.000417 moles of Ar.
03

Calculate molarity

Molarity is defined as moles of solute (Ar) per liter of solvent (water). Considering the water is saturated with air, 1 L of water will contain 0.000417 moles of Ar. Hence, molarity of Ar in air-saturated water at these conditions would be 0.000417 M.

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Most popular questions from this chapter

A saturated solution prepared at \(70^{\circ} \mathrm{C}\) contains \(32.0 \mathrm{g}\) CuSO \(_{4}\) per 100.0 g solution. A 335 g sample of this solution is then cooled to \(0^{\circ} \mathrm{C}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) crystallizes out. If the concentration of a saturated solution at \(0^{\circ} \mathrm{C}\) is \(12.5 \mathrm{g} \mathrm{CuSO}_{4} / 100 \mathrm{g}\) soln, what mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) would be obtained? [Hint: Note that the solution composition is stated in terms of \(\mathrm{CuSO}_{4}\) and that the solid that crystallizes is the hydrate \(\left.\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} .\right]\)

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A \(30.00 \%\) -by-mass solution of nitric acid, \(\mathrm{HNO}_{3},\) in water has a density of \(1.18 \mathrm{g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). What is the molarity of \(\mathrm{HNO}_{3}\) in this solution?

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