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Determine how many grams of each of the following solutes would be needed to make \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.100-M\) solution: (a) cesium iodide (CsI), (b) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right),(\mathrm{c})\) sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right),\) (d) potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right),\) (e) potassium permanganate \(\left(\mathrm{KMnO}_{A}\right)\)

Short Answer

Expert verified
(a) 6.495 g, (b) 2.452 g, (c) 2.650 g, (d) 7.355 g, (e) 3.951 g

Step by step solution

01

Convert Volume to Liters

The volume of the solution is given as \(2.50 \times 10^{2} \text{ mL}\). We need to convert this to liters by using the conversion factor \(1000 \text{ mL} = 1 \text{ L}\). Thus, the volume in liters is \( \frac{2.50 \times 10^{2}}{1000} = 0.250 \text{ L}\).
02

Calculate Moles of Solute

The concentration of the solution is given as \(0.100 \text{ M}\), which means \(0.100 \text{ moles/L}\). To find the number of moles of solute needed, use the formula: \(\text{Moles of Solute} = \text{Concentration} \times \text{Volume}\). Thus, \(0.100 \text{ moles/L} \times 0.250 \text{ L} = 0.025 \text{ moles of solute}\).
03

Calculate Mass for (a) Cesium Iodide (CsI)

First, calculate the molar mass of CsI. Molar mass of Cs (Cesium) is \(132.91 \text{ g/mol}\) and I (Iodine) is \(126.90 \text{ g/mol}\). Thus, \(\text{Molar Mass of CsI} = 132.91 + 126.90 = 259.81 \text{ g/mol}\). Then, calculate the mass using \(\text{Mass} = \text{Moles} \times \text{Molar Mass}\). So, \(0.025 \text{ moles} \times 259.81 \text{ g/mol} = 6.495 \text{ g}\).
04

Calculate Mass for (b) Sulfuric Acid (Hâ‚‚SOâ‚„)

Calculate the molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\). It equals \(2(1.01) + 32.07 + 4(16.00) = 98.09 \text{ g/mol}\). Then, the mass required is \(0.025 \text{ moles} \times 98.09 \text{ g/mol} = 2.452 \text{ g}\).
05

Calculate Mass for (c) Sodium Carbonate (Na₂CO₃)

Calculate the molar mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). It equals \(2(22.99) + 12.01 + 3(16.00) = 105.99 \text{ g/mol}\). The mass is \(0.025 \text{ moles} \times 105.99 \text{ g/mol} = 2.64975 \text{ g}\).
06

Calculate Mass for (d) Potassium Dichromate (K₂Cr₂O₇)

Calculate the molar mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\). It equals \(2(39.10) + 2(51.9961) + 7(16.00) = 294.18 \text{ g/mol}\). Thus, the mass needed is \(0.025 \text{ moles} \times 294.18 \text{ g/mol} = 7.3545 \text{ g}\).
07

Calculate Mass for (e) Potassium Permanganate (KMnOâ‚„)

Calculate the molar mass of \(\mathrm{KMnO}_{4}\). It equals \(39.10 + 54.94 + 4(16.00) = 158.04 \text{ g/mol}\). Therefore, the mass required is \(0.025 \text{ moles} \times 158.04 \text{ g/mol} = 3.951 \text{ g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Molar mass is an essential concept when working with solutions. It is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). To calculate the molar mass, you sum the atomic masses of all the atoms present in the molecule. Atomic masses are usually found on the periodic table.
For instance, to find the molar mass of cesium iodide (CsI), you add the atomic mass of cesium (Cs), which is 132.91 g/mol, to that of iodine (I) which is 126.90 g/mol. The resulting molar mass is 259.81 g/mol.
This method applies to each substance in the exercise. By calculating the molar masses, you can then determine the mass needed for a specific number of moles.
Solution Preparation
When preparing a chemical solution, understanding the desired concentration and volume is critical. A solution's concentration is often expressed in molarity (M), which is the number of moles of solute per liter of solution.
To prepare a solution, you need to determine how many moles of the solute you require, based on the molarity and volume in liters. Multiply the concentration (molarity) by the volume of the solution in liters to get the total moles needed.
In the given exercise, you are asked to make a 0.100 M solution with a total volume of 250 mL, or 0.250 L. Hence, the number of moles is found by multiplying the molarity 0.100 M by the volume 0.250 L, resulting in 0.025 moles of solute.
Chemical Concentration
Chemical concentration measures how much solute is present in a given quantity of solvent or solution. Molarity is a common measure of concentration in chemistry, represented as moles per liter (mol/L).
To prepare a solution with a specific molarity, it's crucial to understand the relationship between moles, volume, and concentration. Knowing the molar mass helps convert between moles and grams, which is often necessary when physically measuring out solutes.
This exercise emphasizes the importance of calculating concentration accurately to ensure the correct mass of each solute is used to produce a solution at the desired molarity. Without precision in concentration, experiments and applications could lead to inaccurate results.
Stoichiometry
Stoichiometry is a fundamental aspect of chemistry that involves quantifying reactants and products in chemical reactions. It relies heavily on the concept of moles and balances equations to ensure the conservation of mass.
While this exercise doesn't delve into reactions, stoichiometric principles are still applied in calculating the mass of solutes needed based on their molarity and volume. Stoichiometry allows chemists to understand the proportions of substances and ensure that reactions proceed with the correct amounts of ingredients.
Though our focus here is on solution preparation, knowing how to use stoichiometry effectively aids in making precise calculations, ensuring experiments and solution preparations are both accurate and effective.

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Most popular questions from this chapter

Classify the following redox reactions as combination. decomposition, or displacement: (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (b) \(\mathrm{Mg}+2 \mathrm{AgNO}_{3} \longrightarrow \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{HBr}\)

Describe in each case how you would separate the cations or anions in the following aqueous solutions: (a) \(\mathrm{NaNO}_{3}\) and \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{K} \mathrm{NO}_{3},\) (c) \(\mathrm{KBr}\) and \(\mathrm{KNO}_{3},\) (d) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KNO}_{3},\) (e) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaNO}_{3}\)

Hydrogen halides \((\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}, \mathrm{HI})\) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and \(\underline{H C l}\) can be generated by combining \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that \(\mathrm{HBr}\) and HI cannot be prepared similarly, that is, by combining NaBr and NaI with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) \((\mathrm{c}) \mathrm{HBr}\) can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

A volume of \(35.2 \mathrm{~mL}\) of a \(1.66 \mathrm{M} \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of a \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

A useful application of oxalic acid is the removal of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) from, say, bathtub rings according to the reaction $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow_{2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{H}^{+}(a q)} $$ Calculate the number of grams of rust that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.100-M\) solution of oxalic acid.

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