Answer :
To determine how many grams of ammonia should be added to 3.62 liters of water to produce a 1.57 molality (m) solution, follow these steps:
1. Understand Molality: Molality is the number of moles of solute (in this case, ammonia) per kilogram of solvent (water).
2. Convert Water Volume to Mass:
- Given: 3.62 liters of water.
- Since the density of water is approximately 1 kg/L, 3.62 liters of water is 3.62 kg.
3. Calculate Moles of Ammonia Needed:
- Given: Desired molality is 1.57 m.
- Using the formula for molality:
[tex]\[
\text{moles of ammonia} = \text{molality} \times \text{mass of water in kg}
\][/tex]
- Substitute the known values:
[tex]\[
\text{moles of ammonia} = 1.57 \times 3.62 = 5.6834 \text{ moles}
\][/tex]
4. Convert Moles of Ammonia to Grams:
- Molar mass of ammonia (NH₃) is approximately 17.03 g/mol.
- Use the formula:
[tex]\[
\text{grams of ammonia} = \text{moles of ammonia} \times \text{molar mass of ammonia}
\][/tex]
- Substitute the known values:
[tex]\[
\text{grams of ammonia} = 5.6834 \times 17.03 = 96.79 \text{ grams}
\][/tex]
Therefore, you should add 96.79 grams of ammonia to 3.62 liters of water to create a 1.57 molal solution.
1. Understand Molality: Molality is the number of moles of solute (in this case, ammonia) per kilogram of solvent (water).
2. Convert Water Volume to Mass:
- Given: 3.62 liters of water.
- Since the density of water is approximately 1 kg/L, 3.62 liters of water is 3.62 kg.
3. Calculate Moles of Ammonia Needed:
- Given: Desired molality is 1.57 m.
- Using the formula for molality:
[tex]\[
\text{moles of ammonia} = \text{molality} \times \text{mass of water in kg}
\][/tex]
- Substitute the known values:
[tex]\[
\text{moles of ammonia} = 1.57 \times 3.62 = 5.6834 \text{ moles}
\][/tex]
4. Convert Moles of Ammonia to Grams:
- Molar mass of ammonia (NH₃) is approximately 17.03 g/mol.
- Use the formula:
[tex]\[
\text{grams of ammonia} = \text{moles of ammonia} \times \text{molar mass of ammonia}
\][/tex]
- Substitute the known values:
[tex]\[
\text{grams of ammonia} = 5.6834 \times 17.03 = 96.79 \text{ grams}
\][/tex]
Therefore, you should add 96.79 grams of ammonia to 3.62 liters of water to create a 1.57 molal solution.
