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Chapter 4: Problem 139

To obtain a solution that is \(1.00 \mathrm{M} \mathrm{NaNO}_{3}\), you should prepare (a) 1.00 L of aqueous solution containing \(100 \mathrm{g} \mathrm{NaNO}_{3} ;\) (b) \(1 \mathrm{kg}\) of aqueous solution containing \(85.0 \mathrm{g} \mathrm{NaNO}_{3} ;(\mathrm{c}) 5.00 \mathrm{L}\) of aqueous solution containing \(425 \mathrm{g} \mathrm{NaNO}_{3} ;(\mathrm{d})\) an aqueous solution containing \(8.5 \mathrm{mg} \mathrm{NaNO}_{3} / \mathrm{mL}\).

Short Answer

Expert verified
The correct solution which gives 1.00 M of \( NaNO_3 \) is not provided in the options.

Step by step solution

01

- Conversion of Quantity of Solute to Moles

Using the molar mass of \( NaNO_3 \), which is 85 g/mol, convert the given quantities of \( NaNO_3 \) in each solution to moles. This is done by dividing the quantity by the molar mass.
02

- Implementation of the Molarity Concept

Apply the concept of molarity, which is number of moles of solute per liter of the solution. The quantity of the solution is given in different units (L, kg and mg/mL), it's crucial to convert these quantities to liters, because molarity is in moles per liter.
03

- Calculation and Comparison with the Desired Molarity

Caculate the molarity for each solution and compare with the desired molarity which is 1.00 M.
04

- Selection of the Appropriate Solution

The solution which gives the desired molarity is chosen as the correct answer. If none of the given solutions give the desired molarity, then the solution that gives a value closest to the desired molarity may be considered.

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Most popular questions from this chapter

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

Iron ore is impure \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) When \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is heated with an excess of carbon (coke), metallic iron and carbon monoxide gas are produced. From a sample of ore weighing \(938 \mathrm{kg}, 523 \mathrm{kg}\) of pure iron is obtained. What is the mass percent \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) by mass, in the ore sample, assuming that none of the impurities contain Fe?

Given a \(0.250 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution, describe how you would prepare a solution that is \(0.0125 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{CrO}_{4} .\) That is, what combination(s) of pipet and volumetric flask would you use? Typical sizes of volumetric flasks found in a general chemistry laboratory are \(100.0,250.0,500.0,\) and \(1000.0 \mathrm{mL},\) and typical sizes of volumetric pipets are 1.00,5.00,10.00 \(25.00,\) and \(50.00 \mathrm{mL}\)

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?

In the reaction of \(2.00 \mathrm{mol} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\), \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. $$ \mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl} $$ (a) The theoretical yield is \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (b) The theoretical yield is \(1.00 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (c) The theoretical yield depends on how large an excess of HF is used. (d) The percent yield is \(85 \%\)
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