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The concentration of commercially available concentrated nitric acid is 70.0 percent by mass, or \(15.9 M .\) Calculate the density and the molality of the solution.

Short Answer

Expert verified
The density is 1.43037 g/mL, and the molality is 37.06 mol/kg.

Step by step solution

01

Understand Definitions

To solve this problem, recall that molarity (M) is defined as the number of moles of solute per liter of solution. Molality is defined as the number of moles of solute per kilogram of solvent. Density is the mass per unit volume of a solution.
02

Define Given Values

We know the solution is 70.0% by mass, it has a molarity (\(M\)) of 15.9, and we need to find its density and molality. Assume 1 liter of solution for easier calculations.
03

Calculate Moles of Solute

Calculate the moles of nitric acid (\(HNO_3\)). Using the molarity, we can find the moles:\[ Moles = M \, \times \, Volume = 15.9 \times 1 = 15.9 \, \text{moles}\]
04

Calculate Mass of Solute

Next, calculate the mass of nitric acid in this 1-liter solution. The molar mass of \(HNO_3\) is 63.01 g/mol, so:\[\text{Mass of } HNO_3 = 15.9 \, \text{moles} \times 63.01 \, \text{g/mol} = 1001.26 \, \text{g}\]
05

Calculate Mass of Solution using Percentage

Since the solution is 70% by mass nitric acid, \(0.7 = \frac{1001.26}{\text{Total Mass}}\). Solving for the total mass, \(\text{Total Mass} = \frac{1001.26}{0.7} = 1430.37 \, \text{g}\).
06

Determine Density of Solution

Density is given by mass per unit volume.\[\text{Density} = \frac{\text{Total Mass}}{\text{Volume}} = \frac{1430.37 \, \text{g}}{1 \, \text{L}} = 1.43037 \, \text{g/mL} = 1.43037 \, \text{kg/L}\]
07

Calculate Mass of Solvent

Determine the mass of the solvent (water) in the solution: \(\text{Total Mass} - \text{Mass of } HNO_3 = 1430.37 - 1001.26 = 429.11 \, \text{g}= 0.42911 \, \text{kg}\).
08

Calculate Molality

Molality is calculated as moles of solute per kilogram of solvent.\[ m = \frac{15.9 \, \text{moles}}{0.42911 \, \text{kg}} = 37.06 \, \text{mol/kg}\]

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Key Concepts

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

Molarity
Molarity is the measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. It is denoted by the symbol "M" and is vital for understanding how "crowded" the solution is with molecules of the solute. When we say a solution is "15.9 M," it simply means there are 15.9 moles of the solute dissolved in each liter of the solution. This makes it easy to compare different solutions and predict how they might react. To calculate molarity, use the formula: \[ \text{Molarity} = \frac{\text{Moles of Solute}}{\text{Liters of Solution}}\] If you know the molarity and want to find the mass of the solute in the solution, multiply the molarity by the volume of the solution in liters and the molar mass of the solute. Understanding molarity is essential when you need to prepare solutions of a specific concentration or when analyzing reactions in a lab.
Molality
Molality measures the concentration of a solution in terms of moles of solute per kilogram of solvent. Unlike molarity, it does not change with temperature because mass is not affected by temperature fluctuations, making it useful for reactions involving temperature change. The symbol for molality is "m."To find molality, use the formula:\[ \text{Molality} = \frac{\text{Moles of Solute}}{\text{Kilograms of Solvent}}\] An example is if you have 15.9 moles of solute in 0.42911 kg of solvent, the molality would be calculated as approximately 37.06 mol/kg. Remember that molality specifically focuses on the solvent mass, not the total mass of the solution, which can sometimes trip people up.
Density Calculation
Density is a measure of how compacted the mass of a substance is within a given volume. For liquids like our nitric acid solution, density is typically expressed in grams per milliliter (g/mL) or kilograms per liter (kg/L). To find the density of our solution, you'd use the formula:\[\text{Density} = \frac{\text{Mass of Solution}}{\text{Volume of Solution}}\] In our exercise, for a 70% by mass nitric acid solution, the calculated density is approximately 1.43037 g/mL. Knowing the density allows you to interconvert between mass and volume, crucial for lab work where precise measurements are necessary. Also, understanding density can help determine how substances will layer or mix when combined.
Percentage by Mass
The percentage by mass of a solution tells you the proportion of the solute mass to the total solution mass, expressed as a percentage. It is calculated using:\[\text{Percentage by Mass} = \left(\frac{\text{Mass of Solute}}{\text{Total Mass of Solution}}\right) \times 100\] In our example, the solution contains 70% nitric acid by mass. This means for every 100 grams of the solution, 70 grams are nitric acid. Percentage by mass is particularly useful when preparing commercial solutions, which are often described in this way to indicate strength and how concentrated they are. This measure can help you determine the mass of solute needed to achieve a specific solution concentration, ensuring accuracy in chemical formulations.

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Most popular questions from this chapter

The concentrated sulfuric acid we use in the laboratory is 98.0 percent \(\mathrm{H}_{2} \mathrm{SO}_{4}\) by mass. Calculate the molality and molarity of the acid solution. The density of the solution is \(1.83 \mathrm{~g} / \mathrm{mL}\)

What is the osmotic pressure (in atm) of a \(1.57-M\) aqueous solution of urea \(\left[\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]\) at \(27.0^{\circ} \mathrm{C} ?\)

The vapor pressure of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) at \(20^{\circ} \mathrm{C}\) is \(44 \mathrm{mmHg},\) and the vapor pressure of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) at the same temperature is \(94 \mathrm{mmHg} .\) A mixture of \(30.0 \mathrm{~g}\) of methanol and \(45.0 \mathrm{~g}\) of ethanol is prepared (and can be assumed to behave as an ideal solution). (a) Calculate the vapor pressure of methanol and ethanol above this solution at \(20^{\circ} \mathrm{C}\). (b) Calculate the mole fraction of methanol and ethanol in the vapor above this solution at \(20^{\circ} \mathrm{C}\). (c) Suggest a method for separating the two components of the solution.

Acetic acid is a polar molecule and can form hydrogen bonds with water molecules. Therefore, it has a high solubility in water. Yet acetic acid is also soluble in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\), a nonpolar solvent that lacks the ability to form hydrogen bonds. A solution of \(3.8 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) in \(80 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{6}\) has a freezing point of \(3.5^{\circ} \mathrm{C}\). Calculate the molar mass of the solute, and suggest what its structure might be. (Hint: Acetic acid molecules can form hydrogen bonds between themselves.)

Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains 80.78 percent \(\mathrm{C}, 13.56\) percent \(\mathrm{H},\) and 5.66 percent \(\mathrm{O}\). A solution of \(1.00 \mathrm{~g}\) of this pheromone in \(8.50 \mathrm{~g}\) of benzene freezes at \(3.37^{\circ} \mathrm{C}\). What are the molecular formula and molar mass of the compound? (The normal freezing point of pure benzene is \(\left.5.50^{\circ} \mathrm{C} .\right)\)

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