Problem 65 - Chemistry The Science in Context

Question

What mass of the following solutions contains 0.100 mol of solute? (a) \(0.135 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3} ;\) (b) \(3.92 \mathrm{m}\) ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} ;(\mathrm{c}) 1.07 \mathrm{m} \mathrm{CaCl}_{2}\)

Step-by-Step Solution

Answer

Question: Calculate the mass of each solution containing 0.100 mol of solute for the following molalities: (a) 0.135 m NH4NO3, (b) 3.92 m ethylene glycol (HOCH2CH2OH), and (c) 1.07 m CaCl2. Answer: (a) 0.135 m NH4NO3 solution: 749.004 g (b) 3.92 m ethylene glycol solution: 31.707 g (c) 1.07 m CaCl2 solution: 104.598 g

Step 1: Calculate the mass of the solvent for each solution

In order to find the mass of each solution, we need to first find the mass of the solvent required to obtain 0.100 mol of the solute in the given concentration (molality). The formula to calculate molality is: \(m = \frac{moles\_of\_solute}{mass\_of\_solvent\_in\_kg}\) Rearranging this equation to find mass of solvent, we get: \(mass\_of\_solvent\_in\_kg = \frac{moles\_of\_solute}{m}\) Calculate the mass of the solvent for each solution: (a) Mass of solvent for 0.135 m NH4NO3: \(mass\_of\_solvent\_in\_kg = \frac{0.100\,mol}{0.135\,m} = 0.741\,kg\) (b) Mass of solvent for 3.92 m ethylene glycol (HOCH2CH2OH): \(mass\_of\_solvent\_in\_kg = \frac{0.100\,mol}{3.92\,m} = 0.0255\,kg\) (c) Mass of solvent for 1.07 m CaCl2: \(mass\_of\_solvent\_in\_kg = \frac{0.100\,mol}{1.07\,m} = 0.0935\,kg\)

Step 2: Calculate the mass of the solute for each solution

Next, find the mass of 0.100 mol of each solute by using their molar masses: (a) Mass of 0.100 mol NH4NO3: \(0.100\,mol \times 80.04\,\frac{g}{mol} = 8.004\,g\) (b) Mass of 0.100 mol ethylene glycol (HOCH2CH2OH): \(0.100\,mol \times 62.07\,\frac{g}{mol} = 6.207\,g\) (c) Mass of 0.100 mol CaCl2: \(0.100\,mol \times 110.98\,\frac{g}{mol} = 11.098\,g\)

Step 3: Calculate the mass of each solution

Finally, add the mass of the solute and the mass of the solvent for each solution to find the total mass of each solution: (a) Mass of 0.135 m NH4NO3 solution: \(8.004\,g \ +\ 741\,g \ =\ 749.004\,g\) (b) Mass of 3.92 m ethylene glycol (HOCH2CH2OH) solution: \(6.207\,g \ +\ 25.5\,g \ =\ 31.707\,g\) (c) Mass of 1.07 m CaCl2 solution: \(11.098\,g \ +\ 93.5\,g \ =\ 104.598\,g\) The masses of each solution containing 0.100 mol of solute are: (a) 0.135 m NH4NO3 solution: 749.004 g (b) 3.92 m ethylene glycol solution: 31.707 g (c) 1.07 m CaCl2 solution: 104.598 g

Key Concepts

Moles of Solute

The concept of moles is fundamentally important in chemistry. Moles measure the amount of substance. When faced with a task such as determining the mass of a solution containing a specific number of moles of solute, it's essential to understand this concept.

• **Moles**: A mole (abbreviated as mol) is a standard scientific unit for measuring large quantities of atoms or molecules.
• **Role in calculations**: Knowing how many moles of a solute are present helps you calculate other properties like mass and concentration.
• **In calculations**: The mole of a solute can be used in formulas such as the molality equation, which ties in the number of moles of solute to the mass of the solvent.

In the given exercise, you encountered 0.100 moles as a fixed quantity for each solute. This is the reference point for figuring out how much solvent is required for solutions of different concentrations, marked by their molality.

Mass of Solvent

Understanding the mass of the solvent is crucial when calculating the properties of a solution, such as its molality. Molality is defined as the number of moles of solute per kilogram of solvent.

• **Mass of Solvent**: This refers to the weight of the solvent only, separate from the solute mixed into the solution.
• **Calculation**: To find this, use the formula derived from molality: \[\text{mass\_of\_solvent\_in\_kg} = \frac{\text{moles\_of\_solute}}{m}\]where \(m\) is the given molality of the solution.
• **Importance in Solution**: Knowing this helps determine the proportion of solvent needed to achieve the desired concentration.

In the exercise, once you know the moles of solute and the molality, you simply rearrange and apply the formula to find how much solvent you'd need for each scenario.

Mass of Solution

The total mass of a solution is the combined mass of both the solute and solvent. Calculating this allows a deeper understanding of the composition and behavior of the solution.

• **Mass of Solution**: This is the sum of the mass of the solute and the mass of the solvent.
• **Calculation**: After determining the mass of solute and solvent separately, they are added together to find the solution mass.
• **Relevance**: Knowing the total mass allows chemists to work with the solution in practical ways, such as when mixing or when reactions are performed in solution state.

In the practice problem, you added the previously computed masses of the solute and solvent to obtain the final solution mass. This result tells you how much the entire solution weighs for the specified amount of solute.