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Chapter 11: Q37 E (page 648)

What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% \({\bf{HN}}{{\bf{O}}_{\bf{3}}}\)by mass)?

  1. Outline the steps necessary to answer the question.
  2. Answer the question.

Short Answer

Expert verified

The molality Of Nitric Acid, \({\bf{HN}}{{\bf{O}}_{\bf{3}}}\,{\bf{is}}\,{\bf{33}}{\bf{.7}}\,{\bf{m}}\)

Step by step solution

01

Molality

Molality is also defined because the number of moles of the solute(n) present in \(1.0\,kg\) of solvent.

\(Molality\left( m \right) = \frac{{{n_{solute}}}}{{Mass{_{solvent}}\,in\,kg}}\)

02

Subpart (a) 

The following steps are followed to calculate molality of the given solution:

i) The number of moles of solute

ii) Substitute the values in molality formula.

\(Molality\left( m \right) = \frac{{{n_{solute}}}}{{Mas{s_{solvent}}\,in\,kg}}\)

03

Subpart (b)

The mass of solute given is 68.0 g.

The molar mass of solute\(HN{O_3}\)is 63.0 g/mol.

The number of moles of solute\(HN{O_3}\)is:

\(\begin{align}{n_{HN{O_3}}} &= \frac{{68.0\,g}}{{63.0\,g/mol}}\\ &= 1.08\,mol\end{align}\)

The mass of solution is 100.0 g.

Hence, the mass of solvent is:

\(\begin{align}{M_{solvent}} &= {M_{solution}} - {M_{solute}}\\ &= 100.0\,g - 68.0\,g\\ &= 32.0\,g\end{align}\)

The mass of solvent in kg is 0.032 kg.

Molality of the solution is:

\(\begin{align}Molality\left( m \right) &= \frac{{{n_{solute}}}}{{Mas{s_{solvent}}\,in\,kg}}\\ &= \frac{{1.08\,mol}}{{0.032\,kg}}\\ &= 33.7\,mol/kg\end{align}\)

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Most popular questions from this chapter

At 0 °C and 1.00 atm, as much as 0.70 g of \({\bf{O}}_2\)can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of \({\bf{O}}_2\)dissolve in 1 L of water?

When \({\bf{KN}}{{\bf{O}}_{\bf{3}}}\) is dissolved in water, the resulting solution is significantly colder than the water was originally.

  1. Is the dissolution of \({\bf{KN}}{{\bf{O}}_{\bf{3}}}\) an endothermic or an exothermic process?
  2. What conclusions can you draw about the intermolecular attractions involved in the process?
  3. Is the resulting solution an ideal solution?

Calculate the mole fraction of each solute and solvent:

  1. 583 g of\({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\)in 1.50 kg of water—the acid solution used in an automobile battery
  2. 0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
  3. 46.85 g of codeine, \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}\), in 125.5 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\)
  4. 25 g of I2 in 125 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\).

What is the freezing temperature of a solution of 115.0 g of sucrose, \({\bf{C_{12}H_{22}O_{11}}}\), in 350.0 g of water, which freezes at \({0.0^{\bf{o}}}{\bf{C}}\) when pure?

  1. Outline the steps necessary to answer the question.
  2. Answer the question.

The Henry’s law constant for \({\bf{O}}_2\)is \({\bf{1}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{ M at}}{{\bf{m}}^{{\bf{ - 1}}}}\)at 25 °C. What mass of oxygen would be dissolved in a 40 L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of \({\bf{O}}_2\)is 0.21 atm?
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