Answer :
Sure! Let's go through each part of the problem step-by-step to find the molality of each solution.
Molality is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per kilogram of solvent.
The formula to calculate molality ([tex]\(m\)[/tex]) is:
[tex]\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \][/tex]
Let's solve each part:
### (a) Sodium carbonate ([tex]\(Na_2CO_3\)[/tex]) in water
1. Determine moles of sodium carbonate:
- Given mass: 0.590 kg
- Molar mass of [tex]\(Na_2CO_3\)[/tex] = 105.99 g/mol
- Convert mass from kg to g: [tex]\(0.590 \, \text{kg} \times 1000 \, \text{g/kg} = 590 \, \text{g}\)[/tex]
- Moles = [tex]\(\frac{590 \, \text{g}}{105.99 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of water = 10.0 kg
- Molality = [tex]\(\frac{\text{moles of } Na_2CO_3}{10.0 \, \text{kg}}\)[/tex]
The calculated molality is approximately 0.5567 mol/kg.
### (b) Ammonium nitrate ([tex]\(NH_4NO_3\)[/tex]) in water
1. Determine moles of ammonium nitrate:
- Given mass: 135 g
- Molar mass of [tex]\(NH_4NO_3\)[/tex] = 80.04 g/mol
- Moles = [tex]\(\frac{135 \, \text{g}}{80.04 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of water = 285 g = 0.285 kg
- Molality = [tex]\(\frac{\text{moles of } NH_4NO_3}{0.285 \, \text{kg}}\)[/tex]
The calculated molality is approximately 5.9181 mol/kg.
### (c) Chlorine ([tex]\(Cl_2\)[/tex]) in dichloromethane ([tex]\(CH_2Cl_2\)[/tex])
1. Determine moles of chlorine:
- Given mass: 55 g
- Molar mass of [tex]\(Cl_2\)[/tex] = 70.90 g/mol
- Moles = [tex]\(\frac{55 \, \text{g}}{70.90 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of dichloromethane = 205 g = 0.205 kg
- Molality = [tex]\(\frac{\text{moles of } Cl_2}{0.205 \, \text{kg}}\)[/tex]
The calculated molality is approximately 3.7841 mol/kg.
### (d) Tetrahydropyridine ([tex]\(C_5H_9N\)[/tex]) in chloroform ([tex]\(CHCl_3\)[/tex])
1. Determine moles of tetrahydropyridine:
- Given mass: 0.355 g
- Molar mass of [tex]\(C_5H_9N\)[/tex] = 83.13 g/mol
- Moles = [tex]\(\frac{0.355 \, \text{g}}{83.13 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of chloroform = 195 g = 0.195 kg
- Molality = [tex]\(\frac{\text{moles of } C_5H_9N}{0.195 \, \text{kg}}\)[/tex]
The calculated molality is approximately 0.0219 mol/kg.
I hope this explanation helps you understand how to calculate molality in these examples! If you have any questions, feel free to ask!
Molality is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per kilogram of solvent.
The formula to calculate molality ([tex]\(m\)[/tex]) is:
[tex]\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \][/tex]
Let's solve each part:
### (a) Sodium carbonate ([tex]\(Na_2CO_3\)[/tex]) in water
1. Determine moles of sodium carbonate:
- Given mass: 0.590 kg
- Molar mass of [tex]\(Na_2CO_3\)[/tex] = 105.99 g/mol
- Convert mass from kg to g: [tex]\(0.590 \, \text{kg} \times 1000 \, \text{g/kg} = 590 \, \text{g}\)[/tex]
- Moles = [tex]\(\frac{590 \, \text{g}}{105.99 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of water = 10.0 kg
- Molality = [tex]\(\frac{\text{moles of } Na_2CO_3}{10.0 \, \text{kg}}\)[/tex]
The calculated molality is approximately 0.5567 mol/kg.
### (b) Ammonium nitrate ([tex]\(NH_4NO_3\)[/tex]) in water
1. Determine moles of ammonium nitrate:
- Given mass: 135 g
- Molar mass of [tex]\(NH_4NO_3\)[/tex] = 80.04 g/mol
- Moles = [tex]\(\frac{135 \, \text{g}}{80.04 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of water = 285 g = 0.285 kg
- Molality = [tex]\(\frac{\text{moles of } NH_4NO_3}{0.285 \, \text{kg}}\)[/tex]
The calculated molality is approximately 5.9181 mol/kg.
### (c) Chlorine ([tex]\(Cl_2\)[/tex]) in dichloromethane ([tex]\(CH_2Cl_2\)[/tex])
1. Determine moles of chlorine:
- Given mass: 55 g
- Molar mass of [tex]\(Cl_2\)[/tex] = 70.90 g/mol
- Moles = [tex]\(\frac{55 \, \text{g}}{70.90 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of dichloromethane = 205 g = 0.205 kg
- Molality = [tex]\(\frac{\text{moles of } Cl_2}{0.205 \, \text{kg}}\)[/tex]
The calculated molality is approximately 3.7841 mol/kg.
### (d) Tetrahydropyridine ([tex]\(C_5H_9N\)[/tex]) in chloroform ([tex]\(CHCl_3\)[/tex])
1. Determine moles of tetrahydropyridine:
- Given mass: 0.355 g
- Molar mass of [tex]\(C_5H_9N\)[/tex] = 83.13 g/mol
- Moles = [tex]\(\frac{0.355 \, \text{g}}{83.13 \, \text{g/mol}}\)[/tex]
2. Calculate molality:
- Mass of chloroform = 195 g = 0.195 kg
- Molality = [tex]\(\frac{\text{moles of } C_5H_9N}{0.195 \, \text{kg}}\)[/tex]
The calculated molality is approximately 0.0219 mol/kg.
I hope this explanation helps you understand how to calculate molality in these examples! If you have any questions, feel free to ask!
