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Chapter 4: Problem 34

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?

Short Answer

Expert verified
Using 10g of AgNO3, approximately 235.6 mL of a 0.25M AgNO3 solution can be prepared.

Step by step solution

01

Find the moles of AgNO3 in the given mass

To calculate the moles of AgNO3, we need to know its molar mass. The molar mass of AgNO3 is: \(M_{AgNO3} = M_{Ag} + M_{N} + 3 \times M_{O}\) Where: \(M_{Ag} \approx 107.87 g/mol\) \(M_{N} \approx 14.01 g/mol\) \(M_{O} \approx 16.00 g/mol\) Now, we can calculate the moles of AgNO3 using the given mass. moles = mass / molar mass
02

Calculate the molar mass of AgNO3

Using the molar masses of Ag, N, and O, we can find the molar mass of AgNO3: \(M_{AgNO3} = 107.87 + 14.01 + 3 \times 16.00 = 169.88 g/mol\)
03

Calculate the moles of AgNO3

Now, we can find the moles of AgNO3 using the given mass of 10g and the molar mass we calculated in Step 2. moles = mass / molar mass moles = 10g / 169.88 g/mol moles ≈ 0.0589 mol
04

Determine the volume of the 0.25M AgNO3 solution

We have the moles of AgNO3 and the desired molarity, so we can calculate the volume of the solution using the molarity formula: M = moles / volume Rearranging the equation to solve for the volume, we get: volume = moles / M Substituting the values we have for the moles and molarity: volume = 0.0589 mol / 0.25 M volume ≈ 0.2356 L
05

Convert the volume to milliliters (if necessary)

If we need the final volume in milliliters, we can convert it as follows: volume = 0.2356 L × 1000 mL/L volume ≈ 235.6 mL Therefore, using 10g of AgNO3, approximately 235.6 mL of a 0.25M AgNO3 solution can be prepared.

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Most popular questions from this chapter

A solution is prepared by dissolving \(10.8 \mathrm{~g}\) ammonium sulfate in enough water to make \(100.0 \mathrm{~mL}\) of stock solution. A \(10.00\) mL sample of this stock solution is added to \(50.00 \mathrm{~mL}\) of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

What mass of solid \(\mathrm{AgBr}\) is produced when \(100.0 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) is added to \(20.0 \mathrm{~mL}\) of \(1.00 M \mathrm{NaBr} ?\)

What volume of each of the following acids will react completely with \(50.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH} ?\) a. \(0.100 \mathrm{M} \mathrm{HCl}\) b. \(0.150 \mathrm{MHNO}_{3}\) c. \(0.200 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) ( 1 acidic hydrogen)

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 \mathrm{M} \mathrm{NaOH}\) from \(1.00 \mathrm{M} \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution
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