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Chapter 13: Problem 29

An aqueous solution is \(34.0 \% \mathrm{H}_{3} \mathrm{PO}_{4}\) by mass and has a density of \(1.209 \mathrm{g} / \mathrm{mL}\). What are the molarity and molality of this solution?

Short Answer

Expert verified
By following the above calculations based on the definitions of molarity and molality, it is possible to solve this exercise. Molarity and molality are important concentrations units in chemistry, and give insights into the properties of a solution, such as boiling point, freezing point, and osmotic pressure.

Step by step solution

01

Determine mass of phosphoric acid and solution

Based on the problem, 34.0% of the solution by mass is \( \mathrm{H}_{3} \mathrm{PO}_{4}\). Therefore, in a 100 g sample of the solution, there are 34 g of \( \mathrm{H}_{3} \mathrm{PO}_{4}\). The mass of the solution (100 g) can be converted into volume using the provided density \((1.209 \mathrm{g} / \mathrm{mL})\). The volume V is calculated by dividing the mass by the density: \(V = 100 \mathrm{g}/(1.209 \mathrm{g/mL})\).
02

Calculate moles of phosphoric acid

The moles of solute are next to be calculated. The molar mass of phosphoric acid \( \mathrm{H}_{3} \mathrm{PO}_{4}\) is \(98.00 \mathrm{g/mol}\). Next, compute the number of moles n of phosphoric acid in the 34 g sample, \(n = 34 \mathrm{g}/(98.00 \mathrm{g/mol})\).
03

Calculate molarity

Molarity (M) is determined as the number of moles of the solute divided by the volume of the solution in liters. With the volume V already calculated to be represented in L and the number of moles n, the molarity is given by: \(M = n/V\).
04

Calculate molality

Molality (m) is given by the number of moles of the solute divided by the mass of the solvent in kilograms. The mass of water in the 100 g solution is \(100 \mathrm{g} - 34 \mathrm{g}\). Converting this to kilograms allows us to calculate the molality as: \(m = n/mass_{water}\)

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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 \(\left.\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} .\right]\)

The term "proof," still used to describe the ethanol content of alcoholic beverages, originated in seventeenthcentury England. A sample of whiskey was poured on gunpowder and set afire. If the gunpowder ignited after the whiskey had burned off, this "proved" that the whiskey had not been watered down. The minimum ethanol content for a positive test was about \(50 \%\) by volume. The \(50 \%\) ethanol solution became known as \(^{\prime \prime} 100\) proof." Thus, an 80 -proof whiskey would be \(40 \% \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) by volume. Listed in the table below are some data for several aqueous solutions of ethanol. With a minimum amount ofcalculation, determine which of the solutions are more than 100 proof. Assume that the density of pure ethanol is \(0.79 \mathrm{g} / \mathrm{mL}\).$$\begin{array}{cl} \hline \text { Molarity of Ethanol, } M & \text { Density of Solution, g/mL } \\\ \hline 4.00 & 0.970 \\\5.00 & 0.963 \\\6.00 & 0.955 \\\7.00 & 0.947 \\\8.00 & 0.936 \\\9.00 & 0.926 \\\10.00 & 0.913 \\\\\hline\end{array}.$$

Explain the observation that all metal nitrates are water soluble but many metal sulfides are not. Among metal sulfides, which would you expect to be most soluble?

What volume of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH},\right.\) density \(=1.12 \mathrm{g} \mathrm{mL}^{-1}\) ) must be added to \(20.0 \mathrm{L}\) of water \(\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} / m\right)\) to produce a solution that freezes at \(-10^{\circ} \mathrm{C} ?\)

Of the following aqueous solutions, the one with the lowest freezing point is (a) \(0.010 \mathrm{mgSO}_{4} ;\) (b) \(0.011 \mathrm{m}\) \(\mathrm{NaCl} ;(\mathrm{c}) 0.018 \mathrm{m} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} ;(\mathrm{d}) 0.0080 \mathrm{m} \mathrm{MgCl}_{2}\).
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