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Chapter 12: Problem 115

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 of the solution is \(1 \frac{kg}{L}\) and the molality of the solution is \(53.0 \frac{moles}{kg}\)

Step by step solution

01

Calculating Mass and Volume

Firstly, we will consider 1 L of solution for calculation. It is given that 70.0 percent by mass is nitric acid. Then, there is \(0.7 kg (700 g)\) of \(HNO_3\) and \(0.3 kg (300 g)\) of water in one liter of solution.
02

Calculating Density

Density is defined as the ratio of mass to volume. The mass of the solution is \(700 g + 300 g = 1000 g = 1 kg\). The density can therefore be calculated as: \(Density = \frac{Mass}{Volume} = \frac{1 kg}{1 L} = 1 \frac{kg}{L}\)
03

Calculating Molarity

Molarity is defined as the no. of moles of solute per litre of solution. As the molarity is given (\(15.9 M\)), the number of moles of \(HNO_3\) in 1L can be calculated as \(15.9 moles\)
04

Calculating Molality

Molality is defined as the no. of moles of solute per kilogram of solvent. As you have already calculated the no. of moles of \(HNO_3\), and we know \('H_2O'\) is 300 g = 0.3 kg, you can simply substitute the values into the formula: \(Molality = \frac{no. of moles of solute}{mass (kg) of solvent} = \frac{15.9 moles}{0.3 kg} = 53.0 \frac{moles}{kg}\)
05

Final Answer

The density of the commercially available nitric acid solution is \(1 \frac{kg}{L}\) and the molality of the solution is \(53.0 \frac{moles}{kg}\)

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Most popular questions from this chapter

Which of the following aqueous solutions has (a) the higher boiling point, (b) the higher freezing point, and (c) the lower vapor pressure: \(0.35 \mathrm{~m}\) \(\mathrm{CaCl}_{2}\) or \(0.90 \mathrm{~m}\) urea? Explain. Assume \(\mathrm{CaCl}_{2}\) to undergo complete dissociation.

What are colligative properties? What is the meaning of the word "colligative" in this context?

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