How many grams of sodium carbonate must be added to 2.50 kg of water in order to lower the freezing point of the resulting solution to 258 Kelvin?
To calculate the amount of sodium carbonate needed to lower the freezing point of the water, we can use the colligative property known as the freezing point depression.
The freezing point depression is given by the formula:
ΔTf = Kf × i × m
Where:
- ΔTf is the change in freezing point (in Kelvin)
- Kf is the cryoscopic constant (molal freezing point depression constant) for water, which is 1.86 °C·kg/mol
- i is the van't Hoff factor, which represents the number of particles the solute dissociates into in solution
- m is the molality of the solution, calculated by dividing the moles of solute by the mass of the solvent (water in this case)
In this problem, we need to find the amount of sodium carbonate required to lower the freezing point to 258 Kelvin. The initial freezing point of pure water is 273 Kelvin, so the change in freezing point (ΔTf) is:
ΔTf = 273 K - 258 K = 15 K
Since sodium carbonate (Na2CO3) is an ionic compound and fully dissociates in solution, the van't Hoff factor (i) for sodium carbonate is 3, as it dissociates into two sodium ions (Na+) and one carbonate ion (CO3^2-).
Now, let's calculate the molality (m). Since we have 2.50 kg of water, we need to convert it to grams:
Mass of water = 2.50 kg × 1000 g/kg = 2500 g
Next, we need to calculate the moles of sodium carbonate (Na2CO3). To do this, we have to consider the molar mass of sodium carbonate, which is:
2 × atomic mass of sodium (Na) + atomic mass of carbon (C) + 3 × atomic mass of oxygen (O)
2 × 22.99 g/mol + 12.01 g/mol + 3 × 16.00 g/mol = 105.99 g/mol
Now, let's calculate the moles of Na2CO3 by dividing the given mass of water by its molar mass:
Moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
To achieve that, we first need to convert the given mass of water to moles:
Moles of water = mass of water / molar mass of water
The molar mass of water is:
2 × atomic mass of hydrogen (H) + atomic mass of oxygen (O) = 2 × 1.01 g/mol + 16.00 g/mol = 18.01 g/mol
Now we can calculate the moles of water:
Moles of water = 2500 g / 18.01 g/mol
With the moles of water calculated, we can now find the moles of sodium carbonate:
Moles of Na2CO3 = Moles of water
Now, to find the molality of the solution, we divide the moles of Na2CO3 by the mass of water (in kg):
m = Moles of Na2CO3 / Mass of water (in kg)
Finally, we can use the freezing point depression formula:
ΔTf = Kf × i × m
Plugging in the values, we can solve for the amount of sodium carbonate (in grams) needed to achieve the desired freezing point depression.