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How many grams of sodium dichromate, \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), should be added to a \(100.0-\mathrm{mL}\) volumetric flask to prepare \(0.025 \mathrm{M} \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) when the flask is filled to the mark with water?

Short Answer

Expert verified
You need 0.655 grams of sodium dichromate.

Step by step solution

01

Calculate the Moles Needed

To find the number of moles of sodium dichromate required, use the formula: \( n = M \times V \), where \( M \) is the molarity (0.025 M) and \( V \) is the volume in liters (0.100 L for 100 mL). \[ n = 0.025 \text{ M }\times 0.100 \text{ L } = 0.0025 \text{ mol}\]
02

Calculate the Molar Mass of Sodium Dichromate

Sodium dichromate \( \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 \) contains 2 sodium (Na), 2 chromium (Cr), and 7 oxygen (O) atoms. Using atomic masses, calculate the molar mass:\(\text{Molar mass of } \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 = (2 \times 22.99) + (2 \times 51.996) + (7 \times 15.999) \ = 45.98 + 103.992 + 111.993 \ = 261.97 \text{ g/mol}\)
03

Calculate the Mass Needed

To find the mass needed, use the formula: \( \text{mass} = n \times \text{molar mass} \), where \( n = 0.0025 \text{ mol} \) and the molar mass is 261.97 g/mol:\[\text{mass} = 0.0025 \text{ mol } \times 261.97 \text{ g/mol } = 0.654925 \text{ g}\]
04

Round and Prepare the Solution

Since the calculated mass is 0.654925 g, round to three significant figures based on precision considerations: \[ \text{Mass needed} \approx 0.655 \text{ g}\]Add 0.655 grams of sodium dichromate to the 100.0 mL volumetric flask and fill it to the mark with water to make a 0.025 M solution.

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Key Concepts

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

Understanding Sodium Dichromate
Sodium dichromate is a bright orange-red crystalline solid that is commonly used in volumetric analysis for redox reactions. It is the sodium salt of dichromic acid and is known for its strong oxidizing properties. This makes it an important chemical for industrial processes and laboratory experiments. When handling sodium dichromate, it is crucial to be cautious as it is a strong oxidizer and can be harmful if ingested or inhaled. Additionally, it is often used in solutions where precise concentration is essential for chemical reactions.
  • Use: Often employed in laboratory settings for its oxidizing capabilities.
  • Appearance: Recognized by its vivid orange crystalline appearance.
  • Safety: Handle with care due to potential health hazards.
Understanding sodium dichromate's role in various chemical processes aids in mastering volumetric analysis and related calculations.
Molarity Calculations Simplified
Molarity is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. In practice, molarity allows chemists to accurately determine how much substance is present in a given volume of liquid, which is vital for reactions requiring precise concentrations.
To calculate molarity, you use the formula:
\[ M = \frac{n}{V} \] where \( n \) is the number of moles of the solute, and \( V \) is the volume of the solution in liters. For example, knowing the molarity of sodium dichromate allows you to calculate how many grams you need to prepare a specific volume of a desired concentration. Here are key concepts related to molarity calculations:
  • Units: Molarity is measured in moles per liter (M).
  • Applications: Used to determine how much solute is required for solutions of specific concentrations.
  • Conversion: Always remember to convert the volume to liters when using the formula.
By mastering molarity calculations, you're equipped to accurately prepare chemical solutions needed for experiments and analyses.
The Basics of Volumetric Analysis
Volumetric analysis is a common laboratory method used to determine the concentration of an unknown substance by reacting it with a standard solution. It involves measuring the precise volumes of liquids, hence the name "volumetric." This method is crucial in many types of chemical analyses, such as titrations. Volumetric analysis often uses a volumetric flask for preparing solutions of known concentration, ensuring precision. When preparing solutions, the following steps can be observed:
  • Volume Measurement: Use precise volumetric tools like pipettes and flasks for accurate results.
  • Standard Solutions: These are solutions with known concentrations and are used to determine the unknown concentrations of other solutions.
  • Application Areas: Widely used for redox reactions, acid-base titrations, and chemical equilibrium studies.
Mastering volumetric analysis allows you to engage in a variety of chemical investigations, providing pivotal knowledge in both academic settings and scientific research.

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Most popular questions from this chapter

Identify each of the following reactions as being a neutralization, precipitation, or oxidation-reduction reaction. a. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) b. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) $$ \mathrm{HgSO}_{4}(s)+2 \mathrm{NaNO}_{3}(a q) $$ c. \(\mathrm{CsOH}(a q)+\mathrm{HClO}_{4}(a q) \longrightarrow\) $$ \mathrm{Cs}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{ClO}_{4}^{-}(a q) $$ d. \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(g)+\mathrm{Na}_{2} \mathrm{~S}(a q) \longrightarrow \mathrm{MgS}(s)+2 \mathrm{NaNO}_{3}(a q)\)

Mercury(II) nitrate is treated with hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), forming a precipitate and a solution. Write the molecular equation and the net ionic equation for the reaction. An acid is formed; is it strong or weak? Name each of the products. If \(81.15 \mathrm{~g}\) of mercury(II) nitrate and \(8.52 \mathrm{~g}\) of hydrogen sulfide are mixed in \(550.0 \mathrm{~g}\) of water to form \(58.16 \mathrm{~g}\) of precipitate, what is the mass of the solution after the reaction?

Obtain the oxidation number for the element noted in each of the following. a. Cr in \(\mathrm{CrO}_{3}\) b. \(\mathrm{Hg}\) in \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) c. Ga in \(\mathrm{Ga}(\mathrm{OH})_{3}\) d. \(\mathrm{P}\) in \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)

Barium carbonate is the source of barium compounds. It is produced in an aqueous precipitation reaction from barium sulfide and sodium carbonate. (Barium sulfide is a soluble compound obtained by heating the mineral barite, which is barium sulfate, with carbon.) What are the molecular equation and net ionic equation for the precipitation reaction? A solution containing \(33.9 \mathrm{~g}\) of barium sulfide requires \(21.2 \mathrm{~g}\) of sodium carbonate to react completely with it, and \(15.6 \mathrm{~g}\) of sodium sulfide is produced in addition to whatever barium carbonate is obtained. How many grams of barium sulfide are required to produce \(5.00\) tons of barium carbonate? (One ton equals 2000 pounds.)

Describe in words how you would do each of the following preparations. Then give the molecular equation for each preparation. a. \(\mathrm{CuCl}_{2}(s)\) from \(\mathrm{CuSO}_{4}(s)\) b. \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(s)\) from \(\mathrm{CaCO}_{3}(s)\) c. \(\mathrm{NaNO}_{3}(s)\) from \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) d. \(\mathrm{MgCl}_{2}(s)\) from \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

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