Question

Calculate the molality of each of the following solutions:

1. 583 g of \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}\) 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, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\).

Short answer

1. The molality of \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}\)is 4 m.
2. The molality of NaCl is 0.14 m.
3. The molality pf Codeine is 1.2 m.
4. The molality of iodine is 0..8 m.

Step-by-step solution

Molality

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

Subpart (a) 

Mass of \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}{\bf{ = 583g}}\)

Molar Mass of \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}{\bf{ = 98 g/mol}}\)

Number of Moles of sulfuric acid

\(\begin{align}{H_2}S{O_4} &= \frac{{583\,g}}{{98\,g/mol}}\\ &= 5.95\end{align}\)

Mass of Water = 1500g

\({\bf{Molality = }}\frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{Moles}}\,{\bf{of}}\,{\bf{solute}}}}{{{\bf{Total}}\,{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent}}\,{\bf{(kg)}}}}\)

Molality of Sulfuric acid

\(\begin{align}{H_2}S{O_4} &= \frac{{5.95\,mol}}{{1500\,g}} \times 1000\\ &= 4\,m\end{align}\)

Subpart (b) 

Mass of \({\bf{NaCl = 0}}{\bf{.86g}}\)

Molar Mass of \({\bf{NaCl = }}58.5{\bf{g/mol}}\)

Number of Moles of NaCl

\(\begin{align}NaCl &= \frac{{0.86\,g}}{{58.5\,g/mol}}\\ &= 0.0147\,mol\end{align}\)

Mass of Water = 102g

\({\bf{Molality = }}\frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{Moles}}\,{\bf{of}}\,{\bf{solute}}}}{{{\bf{Total}}\,{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent}}\,{\bf{(kg)}}}}\)

Molality of NaCl

\(\begin{align}NaCl &= \frac{{0.0147}}{{102}} \times 1000\\ &= 0.14m\end{align}\)

Subpart (c)

Mass of Codeine= 46.85g

Molar Mass of Codeine = 299.4 g/mole

Number of Moles of codeine

\(\begin{align}Codeine &= \frac{{46.85\,g}}{{299.4\,g/mol}}\\ &= 0.156\,mol\end{align}\)

Mass of Ethanol = 125.5g

\({\bf{Molality = }}\frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{Moles}}\,{\bf{of}}\,{\bf{solute}}}}{{{\bf{Total}}\,{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent}}\,{\bf{(kg)}}}}\)

Molality of Codeine

\(\begin{align}Codeine &= \frac{{0.156\,mol}}{{125.5\,g}} \times 1000\\ &= 1.2m\end{align}\)

Subpart (d)

Mass of \({{\bf{I}}_{\bf{2}}}{\bf{ = 25g}}\)

Molar Mass of \({{\bf{I}}_{\bf{2}}}{\bf{ = 253}}{\bf{.81g/mol}}\)

Number of Moles of iodine

\(\begin{align}{I_2} &= \frac{{25\,g}}{{253.81\,g/mol}}\\ &= 0.098\,mol\end{align}\)

Mass of Ethanol = 125g

\({\bf{Molality = }}\frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{Moles}}\,{\bf{of}}\,{\bf{solute}}}}{{{\bf{Total}}\,{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent}}\,{\bf{(kg)}}}}\)

Molality of iodine

\(\begin{align}{I_2} &= \frac{{0.098}}{{125}} \times 1000\\ &= 0.8m\end{align}\)