Final answer:
Approximately 72.15 grams of NaCl will produce the same freezing point depression in 1.25 kg of water as 15.4 grams of urea.
Step-by-step explanation:
To calculate the number of grams of NaCl that will produce the same freezing point as 15.4 grams of urea in 1.25 kg of water, we need to use the colligative property of freezing point depression.
The freezing point depression is determined by the molality of the solute:
ΔT = Kf × m
Where ΔT is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solute.
Since we want the same freezing point depression, we can set up the equation:
Kf × m (NaCl) = Kf × m (urea)
We know the molality of urea is:
m (urea) = 15.4 g / 60.06 g·mol-1 / 1.25 kg = 0.205 m
Let's solve for the molality of NaCl:
m (NaCl) = m (urea)
m (NaCl) = 0.205 m
Now, we can calculate the mass of NaCl:
mass (NaCl) = m (NaCl) · M (NaCl) · mass (water)
Given that the mass of the water is 1.25 kg, we can rearrange the equation to solve for the mass of NaCl:
mass (NaCl) = m (NaCl) · M (NaCl) · 1.25 kg / 0.205 m
We can obtain the molar mass of NaCl from its chemical formula: Na = 22.99 g/mol and Cl = 35.45 g/mol, so:
M (NaCl) = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Substituting the values into the equation, we get:
mass (NaCl) = 0.205 m · 58.44 g/mol · 1.25 kg / 0.205 m = 72.15 g
Therefore, approximately 72.15 grams of NaCl will produce the same freezing point depression in 1.25 kg of water as 15.4 grams of urea.