Answer:
1.61 g
Step-by-step explanation:
The freezing point depression of water due to the presence of salt can be calculated using the equation:
ΔTf = Kf × molality
where ΔTf is the freezing point depression, Kf is the freezing point depression constant for water (1.86 °C/m), and molality is the molal concentration of the solution.
Since ocean water freezes at approximately -1.8 °C, the freezing point depression is:
ΔTf = -1.8 °C - (-0.0 °C) = -1.8 °C
The molality of the solution can be calculated as:
molality = moles of solute / mass of solvent (in kg)
Assuming that all the dissolved salt is sodium chloride (NaCl), the molar mass of NaCl is 58.44 g/mol. Therefore, the number of moles of NaCl in the solution is:
moles of NaCl = mass of NaCl / molar mass of NaCl
moles of NaCl = (2.62 kg) × (35.0 g/kg) / 58.44 g/mol
moles of NaCl = 1.57 mol
The mass of water in the solution is:
mass of water = total mass of solution - mass of NaCl
mass of water = (2.62 kg) - (0.035 kg) = 2.585 kg
Therefore, the molality of the solution is:
molality = 1.57 mol / 2.585 kg = 0.607 mol/kg
Using the freezing point depression equation, we can solve for the mass of NaCl:
ΔTf = Kf × molality
-1.8 °C = 1.86 °C/m × 0.607 mol/kg × (1000 g/1 kg)
-1.8 °C = 1122.42 g/mol × x
x = 1.61 g
Therefore, the mass of sodium chloride in a 2.62-kg sample of ocean water is 1.61 g.