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Chapter 4: Problem 44

What volume of \(0.750 \mathrm{M} \mathrm{AgNO}_{3}\) must be diluted with water to prepare \(250.0 \mathrm{mL}\) of \(0.425 \mathrm{M} \mathrm{AgNO}_{3} ?\)

Short Answer

Expert verified
To prepare \(250.0 \, mL\) of \(0.425 \, M \, \mathrm{AgNO}_{3}\), one must dilute 142 mL of \(0.750 \, M \, \mathrm{AgNO}_{3}\) solution.

Step by step solution

01

Identify Known Variables

The initial molarity, \(M_1\) of \(0.750 \, M\), the final volume, \(V_2\) of \(250.0 \, mL\) or \(0.250 \, L\), and the final molarity, \(M_2\) of \(0.425 \, M\), were provided.
02

Use the Dilution Formula to Solve for \(V_1\)

Substitute the known values into the equation, \(M_1 \cdot V_1 = M_2 \cdot V_2\), and solve for \(V_1 = \frac{M_2 \cdot V_2}{M_1} = \frac{0.425 \, M \cdot 0.250 \, L}{0.750 \, M} = 0.142 \, L\).
03

Convert \(V_1\) to Milliliters (mL)

Convert \(V_1\) from liters to milliliters, using the conversion fact that \(1 L = 1000 mL\), thus \(0.142 \, L = 142 \, mL\).

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Most popular questions from this chapter

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?

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Titanium tetrachloride, \(\mathrm{TiCl}_{4}\) is prepared by the reaction below. $$\begin{aligned} &3 \mathrm{TiO}_{2}(\mathrm{s})+4 \mathrm{C}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{TiCl}_{4}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \end{aligned}$$ What is the maximum mass of \(\mathrm{TiCl}_{4}\) that can be obtained from \(35 \mathrm{g} \mathrm{TiO}_{2^{\prime}} 45 \mathrm{g} \mathrm{Cl}_{2^{\prime}}\) and \(11 \mathrm{g} \mathrm{C} ?\)
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