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Calculate the molarity of each of the following solutions: a. \(64.7 \mathrm{g}\) LiCl in \(250.0 \mathrm{mL}\) of solution b. \(29.3 \mathrm{g} \mathrm{NiSO}_{4}\) in \(200.0 \mathrm{mL}\) of solution c. \(50.0 \mathrm{g} \mathrm{KCN}\) in \(500.0 \mathrm{mL}\) of solution d. \(0.155 \mathrm{g} \mathrm{AgNO}_{3}\) in \(100.0 \mathrm{mL}\) of solution

Short Answer

Expert verified
Question: Calculate the molarity of the following solutions: a. 64.7 g LiCl in 250.0 mL of solution b. 29.3 g NiSO4 in 200.0 mL of solution c. 50.0 g KCN in 500.0 mL of solution d. 0.155 g AgNO3 in 100.0 mL of solution Answer: a. 6.1 M b. 0.95 M c. 1.54 M d. 0.0091 M

Step by step solution

01

Convert the mass of solute to moles

To convert the mass of LiCl to moles, we need its molar mass: Molar mass of LiCl = 6.94 (Li) + 35.45 (Cl) = 42.39 g/mol moles of LiCl = \(\frac{64.7 \mathrm{g}}{42.39 \mathrm{g/mol}}\) = 1.5251 mol
02

Convert the volume of the solution to liters

Volume of solution = 250.0 mL = 0.250 L
03

Calculate the molarity

Molarity (M) = \(\frac{1.5251 \mathrm{mol}}{0.250 \mathrm{L}}\) = 6.1 M b. 29.3 g NiSO4 in 200.0 mL of solution
04

Convert the mass of solute to moles

Molar mass of NiSO4 = 58.69 (Ni) + (4 x 16) (S) + (4 x 16) (O) = 154.75 g/mol moles of NiSO4 = \(\frac{29.3 \mathrm{g}}{154.75 \mathrm{g/mol}}\) = 0.1893 mol
05

Convert the volume of the solution to liters

Volume of solution = 200.0 mL = 0.2 L
06

Calculate the molarity

Molarity (M) = \(\frac{0.1893 \mathrm{mol}}{0.2 \mathrm{L}}\) = 0.95 M c. 50.0 g KCN in 500.0 mL of solution
07

Convert the mass of solute to moles

Molar mass of KCN = 39.10 (K) + 12.01 (C) + 14.01 (N) = 65.12 g/mol moles of KCN = \(\frac{50.0 \mathrm{g}}{65.12 \mathrm{g/mol}}\) = 0.7678 mol
08

Convert the volume of the solution to liters

Volume of solution = 500.0 mL = 0.5 L
09

Calculate the molarity

Molarity (M) = \(\frac{0.7678 \mathrm{mol}}{0.5 \mathrm{L}}\) = 1.54 M d. 0.155 g AgNO3 in 100.0 mL of solution
10

Convert the mass of solute to moles

Molar mass of AgNO3 = 107.87 (Ag) + 14.01 (N) + (3 x 16) (O) = 169.87 g/mol moles of AgNO3 = \(\frac{0.155 \mathrm{g}}{169.87 \mathrm{g/mol}}\) = 0.00091 mol
11

Convert the volume of the solution to liters

Volume of solution = 100.0 mL = 0.1 L
12

Calculate the molarity

Molarity (M) = \(\frac{0.00091 \mathrm{mol}}{0.1 \mathrm{L}}\) = 0.0091 M

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Key Concepts

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

Molar Mass
Understanding molar mass is critical in chemistry, particularly when calculating the molarity of a solution. Molar mass is defined as the mass of one mole of a given substance and is measured in grams per mole (g/mol). Each element has a unique molar mass, which you can find on the periodic table. For compounds, you calculate the molar mass by adding together the molar masses of all the atoms in the compound.

For example, the molar mass calculation for lithium chloride (LiCl) combines the molar masses of lithium (Li) and chlorine (Cl) to yield approximately 42.39 g/mol. This value then helps in converting the mass of the compound into moles, which is essential for understanding the compound's role in a reaction or its concentration in a solution.
Moles Conversion
The mole is a fundamental unit in chemistry representing a specific number of particles, typically atoms or molecules. Converting between moles and grams is a common task in chemistry and involves using the molar mass. To convert grams to moles, you divide the mass of the substance by its molar mass.

In the given exercises, such conversions are necessary to find the number of moles of lithium chloride, nickel(II) sulfate (NiSO4), potassium cyanide (KCN), and silver nitrate (AgNO3). This step is integral in determining the solution's molarity, as the number of moles will later be divided by the volume of the solution in liters.
Solution Concentration
In chemistry, concentration refers to the amount of solute present in a given quantity of solvent or solution. Molarity, symbolized as M, is a common measure of solution concentration and is defined as the number of moles of solute per liter of solution.

To calculate molarity, one must first obtain the number of moles of the solute, which is done by moles conversion, and then divide it by the volume of the solution in liters. In our example, we convert the volume from milliliters to liters and use this value in our molarity calculation, which is essential for stoichiometry calculations in solution chemistry and for laboratory preparations of specific solution concentrations.
Stoichiometry
Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products in chemical reactions. The stoichiometry calculations often require knowledge of molar masses, moles conversions, and solution concentrations. It ensures that quantitative relationships between the reactants and products - as derived from the coefficients of a balanced chemical equation - are understood and applied.

The concept of molarity plays a significant role in stoichiometry, as it provides a link between the volume of a solution and the amount of substance present. With a balanced equation and the molarity of solutions, you can determine how many moles of one substance will react with a certain number of moles of another substance, which is the heart of stoichiometric calculations.

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Most popular questions from this chapter

Complete and balance the molecular equations for the precipitation reactions, if any, between the following pairs of reactants, and write the overall and net ionic equations. a. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) b. \(\mathrm{NiCl}_{2}(a q)+\mathrm{NH}_{4} \mathrm{NO}_{3}(a q) \rightarrow\) c. \(\operatorname{Fe} C l_{2}(a q)+N a_{2} S(a q) \rightarrow\) d. \(\operatorname{MgSO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow\)

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