Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate the molarity of the following aqueous solutions: (a) 0.540 \(\mathrm{g}\) of Mg \(\left(\mathrm{NO}_{3}\right)_{2}\) in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{b}) 22.4 \mathrm{gof}\) \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in 125 \(\mathrm{mL}\) of solution, \((\mathrm{c}) 25.0 \mathrm{mL}\) of 3.50 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to 0.250 \(\mathrm{L}\)

Short Answer

Expert verified
The molarities of the given aqueous solutions are approximately: (a) 0.0146 M Mg(NO3)2 (b) 0.682 M LiClO4.3H2O (c) 0.350 M HNO3

Step by step solution

01

(a) Calculate the Molar Mass of Mg(NO3)2

To find the molarity of the Mg(NO3)2 solution, first calculate the molar mass of Mg(NO3)2. The molar mass of Mg(NO3)2 is obtained by adding the atomic masses of its constituent elements: Mg(NO3)2 = Mg + 2 x (N + 3 x O) = 24.31 g/mol (Mg) + 2 x [14.01 g/mol (N) + 3 x 16.00 g/mol (O)] = 24.31 g/mol + 2 x (14.01 g/mol + 48.00 g/mol) = 24.31 g/mol + 2 x 62.01 g/mol = 148.33 g/mol
02

(a) Calculate moles of Mg(NO3)2

Now you can calculate the moles of Mg(NO3)2 by dividing the mass given (0.540 g) by its molar mass: Moles of Mg(NO3)2 = (0.540 g) / (148.33 g/mol) ≈ 0.00364 mol
03

(a) Convert mL to L

Next, convert the volume of the solution from mL to L: 250.0 mL = 250.0 / 1000 = 0.250 L
04

(a) Calculate Molarity of Mg(NO3)2 Solution

Using the molarity formula (M = moles of solute / liters of solution), you can find the molarity of this solution: Molarity of Mg(NO3)2 = (0.00364 mol) / (0.250 L) ≈ 0.0146 M
05

(b) Calculate the Molar Mass of LiClO4.3H2O

First, find the molar mass of LiClO4.3H2O by adding the atomic masses of its constituent elements: LiClO4.3H2O = Li + Cl + 4 x O + 3 x (2 x H2O) = 6.94 g/mol (Li) + 35.45 g/mol (Cl) + 4 x 16.00 g/mol (O) + 3 x (2 x 1.01 g/mol) (H2O) = 6.94 g/mol + 35.45 g/mol + 64.00 g/mol + 3 x 36.03 g/mol = 262.94 g/mol
06

(b) Calculate moles of LiClO4.3H2O

Now, calculate the moles of LiClO4.3H2O by dividing the given mass of LiClO4.3H2O (22.4 g) by its molar mass: Moles of LiClO4.3H2O = (22.4 g) / (262.94 g/mol) = 0.0852 mol
07

(b) Convert mL to L

Next, convert the volume of the LiClO4.3H2O solution from mL to L: 125 mL = 125 / 1000 = 0.125 L
08

(b) Calculate Molarity of LiClO4.3H2O Solution

Now, use the molarity formula to find the molarity of the LiClO4.3H2O solution: Molarity of LiClO4.3H2O = (0.0852 mol) / (0.125 L) ≈ 0.682 M
09

(c) Find Initial Moles of HNO3

For this solution, you are given the initial concentration (3.50 M) and volume (25.0 mL) of HNO3. First, convert the volume to liters: 25.0 mL = 25.0 / 1000 = 0.025 L Now find the initial moles of HNO3: Initial moles of HNO3 = Initial concentration x Initial volume = (3.50 mol/L) x (0.025 L) ≈ 0.0875 mol
10

(c) Calculate Molarity of Diluted HNO3 Solution

Now, calculate the molarity of the diluted solution using the moles of HNO3 and the final volume (0.250 L): Molarity of diluted HNO3 solution = (0.0875 mol) / (0.250 L) ≈ 0.350 M The molarities of the solutions are approximately as follows: (a) 0.0146 M Mg(NO3)2 (b) 0.682 M LiClO4.3H2O (c) 0.350 M HNO3

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molar Mass Determination
Understanding molar mass is essential for many chemistry calculations. It represents the mass of one mole of a substance, expressed in grams per mole (g/mol). To calculate the molar mass of a compound, like Mg(NO3)2 or LiClO4·3H2O from our exercise, you need to sum the atomic masses of each element present in the compound, taking into account their respective quantity in the molecule.

For instance, Mg(NO3)2 includes one magnesium atom, two nitrogen atoms, and six oxygen atoms. By adding the atomic masses of these elements, using values from the periodic table, you determine the molar mass. This step is fundamental, as the molar mass is then used to convert the mass of the compound into moles, a key step in calculating molarity.
Moles and Molar Concentration
The concept of moles is at the heart of chemistry. It's a measure of quantity that allows chemists to count particles like atoms and molecules. One mole contains Avogadro's number of particles, which is approximately 6.022 × 1023. To move from grams to moles, you divide the mass of a substance by its molar mass.

Molar concentration, often simply called molarity and expressed as 'M', is the number of moles of solute (the substance being dissolved) per liter of solution. The formula to calculate molarity is M = moles of solute / liters of solution. In our exercise, once we calculate the number of moles of our substance, we use this formula to find the molarity, allowing us to understand the concentration of the solution.
Solution Dilution
Dilution involves adding more solvent to a solution to decrease its concentration. The amount of solute remains constant; only the total volume changes. It's crucial to understand that the concentration of a solution is inversely related to its volume when diluting.

Using the formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume, you can calculate the molarity of a diluted solution. In the given exercise, the solution of HNO3 was diluted, and by applying this concept, we can determine its new molarity.
Volume to Liters Conversion
In chemistry experiments, it's often necessary to convert volume measurements to liters, as molarity calculations are based on volume in liters. In everyday life, we encounter various volume measurements, such as milliliters, cups, or gallons. However, since one liter is defined as 1000 milliliters, converting milliliters to liters simply involves dividing the volume by 1000.

In our textbook problem, for example, converting 250.0 mL to liters involves dividing by 1000, yielding 0.250 liters. This conversion is vital for the accurate determination of molarity and is a fundamental part of solution-based calculations in chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The solubility of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water is 208 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\) . A solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O}\) in water at \(35^{\circ} \mathrm{C}\) is formed by dissolving 324 \(\mathrm{g}\) in 100 \(\mathrm{g}\) of water. When this solution is slowly cooled to \(15^{\circ} \mathrm{C},\) no precipitate forms. (a) Is the solution that has cooled down to \(15^{\circ}\) Cunsaturated, saturated, or supersaturated? (b) You take a metal spatula and scratch the side of the glass vessel that contains this cooled solution, and crystals start to appear. What has just happened? (c) At equilibrium, what mass of crystals do you expect to form?

Indicate whether each statement is true or false: (a) The higher the temperature, the more soluble most gases are in water. (b) The higher the temperature, the more soluble most ionic solids are in water. (c) As you cool a saturated solution from high temperature to low temperature, solids start to crystallize out of solution if you achieve a supersaturated solution. (d) If you take a saturated solution and raise its temperature, you can (usually) add more solute and make the solution even more concentrated.

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) boils at \(46.30^{\circ} \mathrm{C}\) and has a density of 1.261 \(\mathrm{g} / \mathrm{mL}\) . (a) When 0.250 \(\mathrm{mol}\) of a nondissociating solute is dissolved in 400.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.46^{\circ} \mathrm{C} .\) What is the molal boiling-point-elevation constant for \(\mathrm{CS}_{2}\) ? (b) When 5.39 \(\mathrm{g}\) of a nondissociating unknown is dissolved in 50.0 \(\mathrm{mL}\) of \(\mathrm{CS}_{2},\) the solution boils at \(47.08^{\circ} \mathrm{C} .\) What is the molar mass of the unknown?

Calculate the number of moles of solute present in each of the following solutions: (a) 255 \(\mathrm{mL}\) of \(1.50 \mathrm{MNO}_{3}(a q),\) (b) 50.0 \(\mathrm{mg}\) of an aqueous solution that is 1.50 \(\mathrm{m} \mathrm{NaCl}_{1}\) (c) 75.0 g of an aqueous solution that is 1.50\(\%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

A "canned heat" product used to warm buffet dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin, which has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{50}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin to produce 8 torr of ethanol vapor pressure at \(35^{\circ} \mathrm{C}\) ? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free