Calculation of Normal Boiling Point of Sea Water

Sea water is found to contain 5.85% NaCl and 9.5% MgCl2 by weight of solution. To calculate its normal boiling point, we need to consider the following steps:

Step 1: Calculate the number of moles of NaCl and MgCl2 in the solution.

Given that the weight of the solution is 100 g, the weight of NaCl present in it is 5.85 g. Similarly, the weight of MgCl2 present in it is 9.5 g.

The molar mass of NaCl is 58.44 g/mol, and that of MgCl2 is 95.21 g/mol.

Therefore, the number of moles of NaCl present in the solution is:

5.85 g / 58.44 g/mol = 0.1 mol

Similarly, the number of moles of MgCl2 present in the solution is:

9.5 g / 95.21 g/mol = 0.1 mol

Step 2: Calculate the total number of moles of solute in the solution.

The total number of moles of solute in the solution is:

0.1 mol (NaCl) + 0.1 mol (MgCl2) = 0.2 mol

Step 3: Calculate the molality of the solution.

The weight of the solvent (water) present in the solution is:

100 g - 5.85 g (NaCl) - 9.5 g (MgCl2) = 84.65 g

The molar mass of water is 18.02 g/mol.

Therefore, the number of moles of water present in the solution is:

84.65 g / 18.02 g/mol = 4.695 mol

The molality of the solution is:

0.2 mol / 4.695 kg = 0.0425 mol/kg

Step 4: Calculate the elevation of boiling point.

The elevation of boiling point is given by the equation:

ΔTb = Kb × molality × i

Where ΔTb is the elevation of boiling point, Kb is the boiling point elevation constant for water (0.52 °C/m), molality is the molality of the solution, and i is the van't Hoff factor.

The van't Hoff factor for NaCl is 2 (because NaCl dissociates into Na+ and Cl- ions), and for MgCl2, it is 3 (because MgCl2 dissociates into Mg2+ and 2Cl- ions).

Therefore, the elevation of boiling point is:

ΔTb = 0.52 °C/m × 0.0425 mol/kg × (2 × 0.7 + 3 × 0.5) = 0.344 °C

Step 5: Calculate the normal boiling point of the solution.

The normal boiling point of water is 100 °C.

Therefore, the normal boiling point of the solution is:

Normal boiling point = 100 °C + ΔTb = 100 °C + 0.344 °C = 100.344 °C

Explanation:

Sea water is a mixture of water

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