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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 鈮� 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 鈮� 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 脳 1000 mL/L volume 鈮� 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

The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.

Which of the following solutions of strong electrolytes contains the largest number of ions: \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}, 50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{BaCl}_{2}\), or \(75.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{4} \mathrm{PO}_{4}\) ?

What volume of \(0.0521 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly \(14.20 \mathrm{~mL}\) of \(0.141 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) ? Phosphoric acid contains three acidic hydrogens.

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.0200 \mathrm{~mol}\) of sodium phosphate in \(10.0 \mathrm{~mL}\) of solution b. \(0.300 \mathrm{~mol}\) of barium nitrate in \(600.0 \mathrm{~mL}\) of solution c. \(1.00 \mathrm{~g}\) of potassium chloride in \(0.500 \mathrm{~L}\) of solution d. \(132 \mathrm{~g}\) of ammonium sulfate in \(1.50 \mathrm{~L}\) of solution

Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion d. sulfate ion b. calcium ion e. mercury(I) ion, \(\mathrm{Hg}_{2}^{2+}\) c. iron(III) ion f. silver ion
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