A chemist studying the propertics of photographic emulsions needed to prepare \(25.00 \mathrm{~mL}\). of \(0.155 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) (aq). What mass of silver nitrate must be placed into a \(25.00-\mathrm{mL}\) volumctric flask and dissolved and diluted to the mark with water?

Short Answer

Expert verified
The chemist should weigh out the mass of \(\mathrm{AgNO}_3\) calculated after multiplying the number of moles (which is determined by the product of the molarity and volume) by the molar mass of \(\mathrm{AgNO}_3\), and then dissolve and dilute to the 25.00 mL mark.

Step by step solution

01

Determine the amount of substance needed

Use the molarity formula, which is molarity (M) = moles of solute/liters of solution. Given a molarity of 0.155 M and a volume of 25.00 mL (which is 0.02500 L), calculate the number of moles of \(\mathrm{AgNO}_3\) needed by multiplying the molarity by the volume in liters: moles = molarity \(\times\) volume.
02

Calculate the mass of \(\mathrm{AgNO}_3\) needed

Find the molar mass of \(\mathrm{AgNO}_3\) from the periodic table (Ag = 107.87 g/mol, N = 14.01 g/mol, O = 16.00 g/mol \(\times\) 3). Then calculate the mass by multiplying the number of moles obtained in Step 1 by the molar mass of \(\mathrm{AgNO}_3\): mass = moles \(\times\) molar mass.
03

Weigh out the mass of \(\mathrm{AgNO}_3\)

Use a balance to measure the calculated mass of silver nitrate and place it into the volumetric flask.
04

Dissolve and dilute to the mark

Dissolve the \(\mathrm{AgNO}_3\) in a small amount of distilled water. Once completely dissolved, add more distilled water until the bottom of the meniscus is at the 25.00 mL mark of the flask.

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Key Concepts

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

Molarity Calculation
Understanding molarity is critical in chemistry, as it represents the concentration of a solution. To calculate molarity, we use the formula:
\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \]
When we know the molarity and the volume of solution we want to prepare, we can rearrange this formula to find the number of moles of solute needed.
\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume} \]
In the given exercise, the chemist uses the molarity of 0.155 M and a volume of 25.00 mL, converted to liters (0.02500 L), to find the moles of \( \mathrm{AgNO}_3 \) required. It's essential to ensure these units are consistent for an accurate calculation.
Molar Mass
To convert moles to mass, which is often required in laboratory settings, we must reference the molar mass of the compound. Molar mass, measured in grams per mole (g/mol), sums up the atomic masses of all atoms in the molecule.
For silver nitrate (\(\mathrm{AgNO}_3\)), we calculate the molar mass by adding the atomic masses of silver (Ag), nitrogen (N), and oxygen (O, multiplied by 3 because there are three oxygen atoms):
\[ \text{Molar mass of } \mathrm{AgNO}_3 = (107.87 + 14.01 + (16.00 \times 3)) \frac{\text{g}}{\text{mol}} \]
This value is crucial to convert the moles of \(\mathrm{AgNO}_3\) calculated in the previous step into grams, providing the exact mass needed to prepare the solution.
Volumetric Analysis
Volumetric analysis involves precisely measuring volumes to determine concentrations of solutions. In the chemist's task, a volumetric flask—a specialized piece of glassware providing high accuracy—is used to prepare a known volume of solution.
Accuracy is essential here; thus, the mark on the flask represents the exact volume at which the water reaches when the desired solution concentration is to be achieved. Ensuring the bottom of the meniscus—the curve seen at the top of the liquid—aligns with the mark is a key step in volumetric analysis.
Only after the solute is fully dissolved, the solution is diluted to the volume mark. This precise process allows chemists to create solutions of exact concentrations, which is fundamental in many chemical applications.
Dilution Method
Dilution is a technique used to reduce the concentration of a solution. It involves adding more solvent without adding more solute. In chemistry, water is often used as the solvent. After weighing and dissolving the calculated mass of solute, such as \(\mathrm{AgNO}_3\), into a small amount of water, additional water is added to reach the target volume.
It's vital to add the solvent incrementally and mix thoroughly to ensure uniform concentration throughout the solution. This method is practical and simple when a lower concentration solution is needed from a more concentrated stock solution. By understanding and applying the dilution method correctly, one can quickly prepare solutions of desired molarities for use in a variety of laboratory experiments.

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Most popular questions from this chapter

(a) What mass (in grams) of \(\mathrm{NaOH}\) must be mixed with \(250 \mathrm{~g}\) of water to prepare a \(0.22 \mathrm{~m} \mathrm{NaOH}\) solution? (b) Calculate the amount (in moles) of cthylene glycol, \(\mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{OH}\), that should be added to \(2.0 \mathrm{~kg}\) of water to prepare a \(0.44 \mathrm{~m} \mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{OH}\) (aq) solution. (c) Determine the amount (in moles) of HCl that must be dissolved in \(500 \mathrm{~g}\) water to prepare a \(0,0010 m\) HClaq) solution.

Calculate the molality of the solute in each of the following solutions: (a) \(10.0 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolvcd in \(250 \mathrm{~g}\) water; (b) \(0.48 \mathrm{~mol}\) of \(\mathrm{KOH}\) dissolved in \(50.0 \mathrm{~g}\) water; (c) \(1.94 \mathrm{~g}\) of urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\), dissolved in \(200 \mathrm{~g}\) water.

Calculate the mole fraction of each component in the following solutions: (a) \(2.5 .0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\); (b) \(25.0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\); (c) a glucowe solution that is \(0.10 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\).

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

The sulfuric acid solution that is purchased for a stockroom has a molarity of \(17.8\) M; all sulfuric acid solutions for experiments are prepared by dilution of this stock solution. (a) Determine the volume of \(17.8 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) that must be diluted to \(250 \mathrm{~mL}\) to prepare a \(2.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) solution. (b) An experiment requires a \(0.50 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) solution. The stockroom manager estimates that \(6.0\) L of the acid is needed. What volume of \(17.8 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) must be used for the preparation?

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