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Chapter 12: Problem 39

Calculate the molarity of each of the following: a. \(2.00 \mathrm{~mol}\) of glucose in \(4.00 \mathrm{~L}\) of a glucose solution b. \(4.00 \mathrm{~g}\) of \(\mathrm{KOH}\) in \(2.00 \mathrm{~L}\) of a \(\mathrm{KOH}\) solution c. \(5.85 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(400 . \mathrm{mL}\) of a \(\mathrm{NaCl}\) solution

Short Answer

Expert verified
a. 0.500 M, b. 0.0357 M, c. 0.250 M

Step by step solution

01

- Determine the formula for molarity

Molarity (\text{M}) is calculated using the formula: \[ \text{M} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
02

- Calculate molarity for part (a)

For part (a), there are \(2.00 \text{ mol}\) of glucose in \(4.00 \text{ L}\) of solution. Using the formula: \[ \text{M} = \frac{2.00 \text{ mol}}{4.00 \text{ L}} = 0.500 \text{ M} \]
03

- Convert grams to moles for part (b)

First, determine the molar mass of KOH: \text{K: }39.1 onumber \text{, O: }16.0 onumber \text{, H: }1.0 onumber\text{Total molar mass: }(39.1 + 16.0 + 1.0) = 56.1 \text{ g/mol}Next, convert \(4.00 \text{ g}\) of KOH to moles: \[ \text{moles} = \frac{4.00 \text{ g}}{56.1 \text{ g/mol}} onumber \text{= 0.0713 mol}\]
04

- Calculate molarity for part (b)

Using the moles calculated and the 2.00 L solution volume: \[ \text{M} = \frac{0.0713 \text{ mol}}{2.00 \text{ L}} = 0.0357 \text{ M} \]
05

- Convert grams to moles for part (c)

First, determine the molar mass of NaCl: \text{Na: }23.0 onumber \text{, Cl: }35.5 onumber\text{Total molar mass: }(23.0 + 35.5) = 58.5 \text{ g/mol}Next, convert \(5.85 \text{ g}\) of NaCl to moles: \[ \text{moles} = \frac{5.85 \text{ g}}{58.5 \text{ g/mol}} onumber \text{= 0.100 mol}\]
06

- Convert solution volume to liters for part (c)

Convert the volume from mL to L: \[400 \text{ mL} = 0.400 \text{ L}\]
07

- Calculate molarity for part (c)

Using the moles calculated and the 0.400 L solution volume: \[ \text{M} = \frac{0.100 \text{ mol}}{0.400 \text{ L}} = 0.250 \text{ M} \]

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Key Concepts

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

Molarity Formula
Understanding molarity is key in chemistry. Molarity, represented as \(M\), measures the concentration of a solution. It's calculated using the formula: \(\text{M} = \frac{\text{moles of solute}}{\text{liters of solution}}\). This means molarity is the number of moles of a solute divided by the total volume of the solution in liters. Molarity helps us understand the strength of a solution. For example, a solution with a molarity of 1 M means it has 1 mole of solute per liter of solution. Remember to always use liters when calculating molarity!
Converting Grams to Moles
To use the molarity formula, we often need to convert grams to moles. Moles represent a quantity of a chemical substance. To convert grams to moles, use the molar mass of the substance. The molar mass is found on the periodic table. For instance, the molar mass of KOH is calculated by adding the atomic masses of Potassium (K: 39.1 g/mol), Oxygen (O: 16.0 g/mol), and Hydrogen (H: 1.0 g/mol). Thus, the molar mass of KOH is 56.1 g/mol. To convert grams to moles, divide the mass of the substance by its molar mass. Example: For 4.00 g of KOH: \(\text{moles} = \frac{4.00 \text{ g}}{56.1 \text{ g/mol}} = 0.0713 \text{ mol}\).
Chemical Solutions
A chemical solution is a homogeneous mixture composed of two or more substances. The solute is the substance dissolved, and the solvent is the substance in which the solute is dissolved. In our examples, glucose, KOH, and NaCl are solutes. When calculating molarity, it's important to use the total volume of the solution. For example, in part (a), we have 2.00 mol of glucose in a 4.00 L solution. The total volume includes both the solute and the solvent. Always ensure your volume is in liters to correctly use the molarity formula.
Molar Mass Calculation
Calculating the molar mass of a compound is an essential skill in chemistry. The molar mass is the sum of the atomic masses of all atoms in a molecule. For NaCl, add the atomic masses of Sodium (Na: 23.0 g/mol) and Chlorine (Cl: 35.5 g/mol) to get 58.5 g/mol. Once you have the molar mass, you can convert grams of a substance to moles. For example, converting 5.85 g of NaCl: \(\text{moles} = \frac{5.85 \text{ g}}{58.5 \text{ g/mol}} = 0.100 \text{ moles}\). Using this value, you can proceed to calculate the molarity if you have the volume of the solution.

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Most popular questions from this chapter

In each pair, identify the solution that will have a higher boiling point. Explain. a. \(1.50 \mathrm{~mol}\) of \(\mathrm{LiOH}\) (strong electrolyte) and \(3.00 \mathrm{~mol}\) of \(\mathrm{KOH}\) (strong electrolyte) each in \(1.0 \mathrm{~kg}\) of water b. \(0.40 \mathrm{~mol}\) of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (strong electrolyte) and \(0.40 \mathrm{~mol}\) of \(\mathrm{CsCl}\) (strong electrolyte) each in \(1.0 \mathrm{~kg}\) of water

Calculate the mass percent \((\mathrm{m} / \mathrm{m})\) for the solute in each of the following: a. \(75 \mathrm{~g}\) of \(\mathrm{NaOH}\) in \(325 \mathrm{~g}\) of \(\mathrm{NaOH}\) solution b. \(2.0 \mathrm{~g}\) of \(\mathrm{KOH}\) and \(20.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) c. \(48.5 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in \(250.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution

Classify the solute represented in each of the following equations as a strong, weak, or nonelectrolyte: a. \(\mathrm{K}_{2} \mathrm{SO}_{4}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} 2 \mathrm{~K}^{+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftarrows \stackrel{\mathrm{NH}_{4}^{+}(a q)}+\mathrm{OH}^{-}(a q)\) c. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\)

Answer the following for the reaction: $$ \mathrm{NiCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{NaCl}(a q) $$ a. How many milliliters of a \(0.200 \mathrm{M} \mathrm{NaOH}\) solution are needed to react with \(18.0 \mathrm{~mL}\) of a \(0.500 \mathrm{M} \mathrm{NiCl}_{2}\) solution? b. How many grams of \(\mathrm{Ni}(\mathrm{OH})_{2}\) are produced from the reaction of \(35.0 \mathrm{~mL}\) of a \(1.75 \mathrm{M} \mathrm{NaOH}\) solution and excess \(\mathrm{NiCl}_{2} ?\) c. What is the molarity of \(30.0 \mathrm{~mL}\) of a \(\mathrm{NiCl}_{2}\) solution that reacts completely with \(10.0 \mathrm{~mL}\) of a \(0.250 \mathrm{M} \mathrm{NaOH}\) solution?

A \(0.1 \%(\mathrm{~m} / \mathrm{v})\) albumin solution is separated from a \(2 \%(\mathrm{~m} / \mathrm{v})\) albumin solution by a semipermeable membrane. (Albumin is a colloid.) a. Which compartment has the higher osmotic pressure? b. In which direction will water flow initially? c. In which compartment will the volume level rise?
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