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To obtain a solution that is 1.00MNaNO3, you should prepare (a) 1.00 L of aqueous solution containing 100gNaNO3; (b) 1kg of aqueous solution containing 85.0gNaNO3;(c)5.00L of aqueous solution containing 425gNaNO3;(d) an aqueous solution containing 8.5mgNaNO3/mL.

Short Answer

Expert verified
The correct solution which gives 1.00 M of NaNO3 is not provided in the options.

Step by step solution

01

- Conversion of Quantity of Solute to Moles

Using the molar mass of NaNO3, which is 85 g/mol, convert the given quantities of NaNO3 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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Key Concepts

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

Molarity
Molarity is a key concept in chemistry that measures the concentration of a solution. It is specifically defined as the number of moles of solute per liter of solution. Understanding how to calculate molarity is crucial as it allows chemists to communicate concentrations accurately. The formula to determine molarity is:
  • Molarity ( M ) = Moles of solute / Liters of solution.
This relationship highlights why it’s essential to ensure the volume of the solution is converted to liters, assuring that molarity calculations are consistent and reliable.
When working with molarity, keep in mind that it is always expressed in terms of liters. This requirement underlines the importance of unit conversions in chemistry, which we will explore in another section.
Moles Conversion
Converting grams of a compound to moles is a fundamental skill in chemistry. To convert grams to moles, you'll use the molar mass of the compound, which is the mass of one mole of that substance. The formula to convert grams to moles is:
  • Moles = Mass of substance (g) / Molar mass (g/mol).
In the context of Sodium Nitrate ( NaNO_3 ), if you are given the mass of NaNO_3 in grams, you divide this by its molar mass (85 g/mol) to determine the number of moles. This conversion is a key step in determining the molarity of a solution.
By effectively converting grams to moles, you gain an understanding of the relationship between the amount of substance and its chemical potential in reactions. This is an essential process for calculating the concentration of a solution.
Sodium Nitrate (NaNO3)
Sodium nitrate ( NaNO_3 ) is a common chemical compound with varied applications in industry and laboratories. It is known for its usage in fertilizers, food preservation, and application in manufacturing explosives. Understanding the chemical properties and uses of NaNO_3 is important when preparing solutions.
  • It is a crystalline solid, highly soluble in water.
  • Molar mass: 85 g/mol, crucial for conversion calculations.
The role of Sodium Nitrate in solution preparation is pivotal due to its chemical stability and reactivity. When preparing a solution with a specific molarity, knowing the precise molar mass is necessary for accurate molarity calculations. This ensures the right amount of solute is dissolved, leading to an intended concentration.
Volume Conversion
Volume conversion is an essential task when working with solutions, particularly for determining molarity. Molarity calculations require that the volume of the solution is expressed in liters. Sometimes, you may be given a solution’s volume in milliliters (mL) or kilograms (kg). Here’s how to handle these conversions:
  • To convert milliliters to liters: divide by 1,000 (since 1,000 mL = 1 L).
  • For density-dependent conversions from kilograms: consider the type of solute and solvent used. For instance, assuming the solution has water as a base, know that 1 kg of water is approximately 1 liter.
Volume conversions ensure that all measurements are within the correct units for straightforward calculations. This step is vital to ensure molarity calculations are accurate and effective, allowing chemists to reach desired solution concentrations.

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Most popular questions from this chapter

What are the molarities of the following solutes? (a) aspartic acid (H2C4H5NO4) if 0.405g is dissolved in enough water to make 100.0mL of solution (b) acetone, C3H6O,(d=0.790g/mL) if 35.0mL is dissolved in enough water to make 425mL of solution (c) diethyl ether, (C2H5)2O, if 8.8mg is dissolved in enough water to make 3.00 L of solution

Water and ethanol, CH3CH2OH(l), are miscible, that is, they can be mixed in all proportions. However, when these liquids are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes, and we say that the volumes are not additive. For example, when 50.0mL of water and 50.0mL of CH3CH2OH(l), are mixed at 20C, the total volume of the solution is 96.5mL, not 100.0mL. (The volumes are not additive because the interactions and packing of water molecules are slightly different from the interactions and packing of CH3CH2OH molecules.) Calculate the molarity of CH3CH2OH in a solution prepared by mixing 50.0mL of water and 50.0mL of CH3CH2OH(l) at 20C. At this temperature, the densities of water and ethanol are 0.99821 g/mL and 0.7893g/mL, respectively.

The following set of reactions is to be used as the basis of a method for producing nitric acid, HNO3 Calculate the minimum masses of N2,H2 and O2 required per kilogram of HNO3 N2(g)+3H2(g)2NH3(g)4NH3(g)+5O2(g)4NO(g)+6H2O(g)2NO(g)+O2(g)2NO2(g)3NO2(g)+H2O(l)2HNO3(aq)+NO(g)

An essentially 100% yield is necessary for a chemical reaction used to analyze a compound, but it is almost never expected for a reaction that is used to synthesize a compound. Explain this difference.

How many moles of NO(g) can be produced in the reaction of 3.00molNH3(g) and 4.00molO2(g)? 4NH3(g)+5O2(g)Δ4NO(g)+6H2O(l)

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