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Chapter 13: Problem 43

Calculate the molarity of the following aqueous solutions: (a) \(0.640 \mathrm{~g}\) of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in \(500.0 \mathrm{~mL}\) of solution, (b) \(50.0 \mathrm{~g}\) of \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in $250 \mathrm{~mL}$ of solution, (c) \(125 \mathrm{~mL}\) of \(3.00 \mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to \(1.00 \mathrm{~L}\)

Short Answer

Expert verified
The molarity of the aqueous solutions are: (a) \(0.00862 \, M\) for the \(Mg(NO_3)_2\) solution, (b) \(1.03 \, M\) for the \(LiClO_4 \cdot 3H_2O\) solution, and (c) \(0.375 \, M\) for the diluted \(HNO_3\) solution.

Step by step solution

01

Calculate moles of solute

First, determine the molar mass of \(Mg(NO_3)_2\). The molar mass is the sum of the atomic masses of the elements in the compound. \(Mg(NO_3)_2 = Mg + 2(N + 3O) = 24.31 + 2(14.01+3(16.00))\) Molar mass of \(Mg(NO_3)_2 = 148.33 \, g/mol\) Now, calculate the moles of \(Mg(NO_3)_2\): Moles of \(Mg(NO_3)_2 = \cfrac{0.640 \, g}{148.33 \, g/mol} = 0.00431 \, mol\)
02

Convert volume to liters

Now, convert the given volume of solution from milliliters to liters: $$500.0 \, mL \times \frac{1 \, L}{1000 \, mL} = 0.500 \, L$$
03

Calculate molarity

Now, calculate the molarity by dividing moles of solute by the volume of solution in liters: $$M = \cfrac{0.00431 \, mol}{0.500 \, L} = 0.00862 \, M$$ The molarity of the solution is \(0.00862 \, M\). (b) Calculate the molarity of a solution containing 50.0 g of LiClO4·3H2O in 250 mL of solution
04

Calculate moles of solute

First, determine the molar mass of \(LiClO_4∙3H_2O\): \(LiClO_4∙3H_2O = Li + Cl + 4O + 3(2H + O)\) Molar mass of \(LiClO_4∙3H_2O = 194.10 \, g/mol\) Now, calculate the moles of \(LiClO_4∙3H_2O\): Moles of \(LiClO_4∙3H_2O = \cfrac{50.0 \, g}{194.10 \, g/mol} = 0.257 \, mol\)
05

Convert volume to liters

Now, convert the given volume of solution from milliliters to liters: $$250.0 \, mL \times \frac{1 \, L}{1000 \, mL} = 0.250 \, L$$
06

Calculate molarity

Now, calculate the molarity by dividing moles of solute by the volume of solution in liters: $$M = \cfrac{0.257 \, mol}{0.250 \, L} = 1.03 \, M$$ The molarity of the solution is \(1.03 \, M\). (c) Calculate the molarity of 125 mL of a 3.00 M HNO3 solution diluted to 1.00 L
07

Calculate moles of solute in the initial solution

First, find the moles of \(HNO_3\) in the initial \(125 \, mL\) solution: $$Moles \, of \, HNO_3 = Molarity \times Volume = 3.00 \, M \times 0.125 \, L = 0.375 \, mol$$
08

Calculate molarity in the diluted solution

The moles of solute do not change during the dilution process, so to calculate the molarity in the final diluted solution, divide the moles of solute by the final volume of solution in liters: $$M = \cfrac{0.375 \, mol}{1.00 \, L} = 0.375 \, M$$ The molarity of the diluted solution is \(0.375 \, M\).

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Most popular questions from this chapter

Common laboratory solvents include acetone $\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\(, methanol \)\left(\mathrm{CH}_{3} \mathrm{OH}\right)\(, toluene \)\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\right),$ and water. Which of these is the best solvent for nonpolar solutes?

A sulfuric acid solution containing \(697.6 \mathrm{~g}\) of $\mathrm{H}_{2} \mathrm{SO}_{4}\( per liter of solution has a density of \)1.395 \mathrm{~g} / \mathrm{cm}^{3} .$ Calculate (a) the mass percentage, \((\mathbf{b})\) the mole fraction, (c) the molality, \((\mathbf{d})\) the molarity of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in this solution.

Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing \(0.150 \mathrm{~g}\) of this enzyme in \(210 \mathrm{~mL}\) of solution has an osmotic pressure of \(0.127 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\). What is the molar mass of lysozyme?

Which of the following in each pair is likely to be more soluble in hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}:\) (a) \(\mathrm{CCl}_{4}\) or \(\mathrm{CaCl}_{2}\), (b) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) or glycerol, $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH},\( (c) octanoic acid, \)\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},$ or acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\) ? Explain your answer in each case.

Ascorbic acid (vitamin C, $\left.\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\( is a water-soluble vitamin. A solution containing \)80.5 \mathrm{~g}\( of ascorbic acid dissolved in \)210 \mathrm{~g}$ of water has a density of \(1.22 \mathrm{~g} / \mathrm{mL}\) at \(55^{\circ} \mathrm{C}\). Calculate (a) the mass percentage, (b) the mole fraction, \((\mathbf{c})\) the molality, \((\mathbf{d})\) the molarity of ascorbic acid in this solution.
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