What is the freezing point in of a 0.52 m aqueous solution of naci
fp(water) = o.oo c
kf (water)=1.86 c/m
To calculate the freezing point depression of the solution, we use the formula:
ΔTf = Kf · m
where ΔTf is the freezing point depression, Kf is the freezing point depression constant of water (1.86 °C/m), and m is the molality of the solute (moles of solute per kilogram of solvent).
First, we need to calculate the molality of the NaCl solution. We know that the concentration of the solution is 0.52 m, which means there are 0.52 moles of NaCl per 1 liter of solution. However, we need to know the mass of the solvent (water) to calculate the molality. Assuming the density of the solution is 1 g/mL, we can calculate the mass of 1 liter of solution:
1000 g (1 L) / 1 mL = 1000 g
So, 1 liter of the solution has a mass of 1000 g. Since we know the concentration of the solution in moles per liter, we can calculate the number of moles of NaCl in 1000 g of solution:
0.52 moles / 1 liter * 1 liter / 1000 g * 1000 g = 0.52 moles
Therefore, the molality of the solution is:
m = 0.52 moles / 1 kg = 0.52 m
Now, we can use the formula above to calculate the freezing point depression:
ΔTf = 1.86 °C/m * 0.52 m = 0.9672 °C
Since the freezing point of pure water is 0.00 °C, the freezing point of the NaCl solution is:
0.00 °C - 0.9672 °C = -0.9672 °C
Therefore, the freezing point of a 0.52 m aqueous solution of NaCl is -0.9672 °C.