/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 132 An aqueous solution is \(20.0 \%... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An aqueous solution is \(20.0 \%\) by mass of sodium thiosulfate pentahydrate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O} .\) What is the molarity of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) in this solution at \(20^{\circ} \mathrm{C} ?\) The density of this solution at \(20^{\circ} \mathrm{C}\) is \(1.174 \mathrm{~g} / \mathrm{mL}\)

Short Answer

Expert verified
The molarity of sodium thiosulfate pentahydrate in the solution is approximately 0.951 M.

Step by step solution

01

Understand the problem

We need to find the molarity of a sodium thiosulfate pentahydrate solution. We are given the mass percentage, density of the solution, and we need to find the molarity of the solute by converting the given percentage mass into moles per liter.
02

Calculate the mass of solute in 100 g of solution

Since the solution is 20.0% by mass of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\),in 100 g of solution, there are 20.0 g of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\).
03

Determine the molar mass of the solute

Calculate the molar mass: \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\). Use atomic masses: Na = 23, S = 32, O = 16, H = 1. First, calculate \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \): 2(23) + 2(32) + 3(16) = 158 \text{g/mol}Then, add 5 water molecules: 5[2(1) + 16] = 90 \text{g/mol}Total molar mass is 158 + 90 = 248 \text{g/mol}.
04

Calculate moles of solute in 100 g of solution

Using the mass of the solute and its molar mass:\[\text{Moles of } \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O} = \frac{20.0 \text{ g}}{248 \text{ g/mol}} \approx 0.081 \text{ moles}\]
05

Find the volume of solution in liters

Use the density given to convert 100 g of solution to volume:\[\text{Volume in mL} = \frac{100 \text{ g}}{1.174 \text{ g/mL}} \approx 85.17 \text{ mL}\]Convert to liters:\[85.17 \text{ mL} = 0.08517 \text{ L}\]
06

Calculate the molarity of the solution

Molarity is moles per liter. Use the number of moles from Step 4:\[\text{Molarity} = \frac{0.081 \text{ moles}}{0.08517 \text{ L}} \approx 0.951 \text{ M}\]

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 91Ó°ÊÓ!

Key Concepts

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

Sodium Thiosulfate Pentahydrate
Sodium thiosulfate pentahydrate, written as \( \text{Na}_2 \text{S}_2 \text{O}_3 \cdot 5 \text{H}_2 \text{O} \), is an important chemical compound often used in various applications, including photography and water treatment. It's a hydrated salt, meaning it contains water molecules within its crystalline structure. The term "pentahydrate" indicates that there are five water molecules attached to each formula unit of sodium thiosulfate.
This hydration makes sodium thiosulfate pentahydrate distinct from its anhydrous (water-free) form, as the presence of water molecules can notably affect the compound's physical properties like crystal structure and solubility in water.
Density and Solution Concentration
Density is a key concept when working with solutions, as it helps convert between volume and mass. Density is defined as mass divided by volume, typically expressed in grams per milliliter (g/mL). In the exercise, we're given that the density of the solution is 1.174 g/mL at 20°C. This means 1 milliliter of the solution has a mass of 1.174 grams. Knowing the density is crucial for determining the solution concentration, which refers to how much solute is present in a given volume of solution.
To find concentration by molarity, you often start with the known mass percent and the density to calculate how much solute is present in a particular volume, then convert this into moles using the solute's molar mass.
Molar Mass Calculation
Molar mass is the mass of a substance divided by the amount of substance (moles) and is expressed in grams per mole (g/mol). For sodium thiosulfate pentahydrate, calculating its molar mass involves adding up the atomic masses of all the atoms in its formula.
  • The sodium (Na) content is calculated as \(2 \times 23 = 46\) g/mol.
  • The sulfur (S) content is \(2 \times 32 = 64\) g/mol.
  • The oxygen (O) content is \(3 \times 16 = 48\) g/mol.
Adding these gives the molar mass for \(\text{Na}_2\text{S}_2\text{O}_3\), totaling 158 g/mol.
Additionally, for the five water molecules, \( 5 \times (2 \times 1 + 16) = 90 \) g/mol is added. Therefore, the total molar mass of sodium thiosulfate pentahydrate is 248 g/mol, which is crucial for further moles-per-volume calculations.
Mass Percentage
Mass percentage is a common way of expressing concentration in a solution. It indicates the ratio of the mass of the solute to the total mass of the solution, multiplied by 100 to convert it into a percentage. In this case, a 20% mass percentage means there are 20 grams of sodium thiosulfate pentahydrate for every 100 grams of solution. Understanding mass percentage allows you to determine how much of the solute is contained in any amount of solution you are working with.
This percentage can then be used along with the solution's density to figure out the actual concentration, as mass gives a starting point for converting to moles, necessary for finding molarity.
Aqueous Solutions
An aqueous solution is simply a solution in which water is the solvent. This means the solute is dissolved in water. Aqueous solutions are central to chemistry because many reactions take place in this medium. In the problem, sodium thiosulfate pentahydrate is dissolved in water which forms the aqueous solution.
Understanding aqueous solutions is important because:
  • They allow for chemical reactions to occur in a liquid state.
  • Water, being a universal solvent, supports a wide range of solutes, including ionic and molecular compounds.
  • They are also essential in understanding how various compounds interact at a molecular level.
Hence, calculating molarity in such solutions becomes a fundamental part of evaluating the concentration and understanding the solution's overall properties.

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 osmotic pressure of blood at \(37^{\circ} \mathrm{C}\) is \(7.7 \mathrm{~atm} . \mathrm{A}\) solution that is given intravenously must have the same osmotic pressure as the blood. What should be the molarity of a glucose solution to give an osmotic pressure of \(7.7 \mathrm{~atm}\) at \(37^{\circ} \mathrm{C}\) ?

Consider two hypothetical pure substances, \(\mathrm{AB}(s)\) and \(\mathrm{XY}(s) .\) When equal molar amounts of these substances are placed in separate \(500-\mathrm{mL}\) samples of water, they undergo the following reactions: $$\begin{gathered}\mathrm{AB}(s) \longrightarrow \mathrm{A}^{+}(a q)+\mathrm{B}^{-}(a q) \\ \mathrm{XY}(s) \longrightarrow \mathrm{XY}(a q) \end{gathered}$$ a. Which solution would you expect to have the lower boiling point? Why? b. Would you expect the vapor pressures of the two solutions to be equal? If not, which one would you expect to have the higher vapor pressure? C. Describe a procedure that would make the two solutions have the same boiling point. d. If you took \(250 \mathrm{~mL}\) of the \(\mathrm{AB}(a q)\) solution prepared above, would it have the same boiling point as the original solution? Be sure to explain your answer. e. The container of \(X Y(a q)\) is left out on the bench top for several days, which allows some of the water to evaporate from the solution. How would the melting point of this solution compare to the melting point of the original solution?

An aqueous solution is \(0.797 M\) magnesium chloride. Given that the density of the solution is \(1.108 \mathrm{~g} / \mathrm{mL}\), what is the molality of magnesium chloride in this solution?

Vanillin, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\), occurs naturally in vanilla extract and is used as a flavoring agent. A 39.1-mg sample of vanillin was dissolved in \(168.5 \mathrm{mg}\) of diphenyl ether, \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{O} .\) What is the molality of vanillin in the solution?

Ten grams of the hypothetical ionic compounds \(\mathrm{XZ}\) and YZ are each placed in a separate \(2.0-\mathrm{L}\) beaker of water. \(\mathrm{XZ}\) completely dissolves, whereas YZ is insoluble. The energy of hydration of the \(\mathrm{Y}^{+}\) ion is greater than the \(\mathrm{X}^{+}\) ion. Explain this difference in solubility.

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.