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Calculate the molality of each of the following solutions:

(a) 0.590 kg of sodium carbonate (washing soda), [tex]Na_2CO_3[/tex], in 10.0 kg of water.

(b) 135 g of [tex]NH_4NO_3[/tex] in 285 g of water.

(c) 55 g of [tex]Cl_2[/tex] in 205 g of dichloromethane, [tex]CH_2Cl_2[/tex].

(d) 0.355 g of tetrahydropyridine, [tex]C_5H_9N[/tex], in 195 g of chloroform, [tex]CHCl_3[/tex].

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!

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