If you had 1L of water, how many grams of NaCl will youi need to mix to the water to lower the freezing point by 24 degrees C?
To calculate the amount of NaCl needed to lower the freezing point of water by a certain number of degrees, you can use the equation for the cryoscopic constant, which is the constant that relates the molality of the solute to the change in freezing point. The equation is:
ΔT = Kf × m
Where:
ΔT is the change in freezing point (in degrees Celsius),
Kf is the cryoscopic constant (which is 1.86 °C·kg/mol for water),
m is the molality of the solute (in mol/kg).
In this case, we want to lower the freezing point by 24 degrees Celsius. So the value of ΔT is -24 °C (negative because we are lowering the freezing point). The volume of water is given as 1 liter, which is equivalent to 1 kg because the density of water is approximately 1 g/mL.
Now, we can solve for the molality of the solute (NaCl) using the rearranged equation:
m = ΔT / (Kf × 1)
Plugging in the given values:
m = -24 / (1.86 × 1)
m ≈ -12.9 mol/kg
Since molality (m) is the number of moles of solute per kilogram of solvent, we can determine the number of moles of NaCl needed to achieve this molality by multiplying the molality by the mass of the water in kilograms:
moles of NaCl = -12.9 mol/kg × 1 kg
moles of NaCl ≈ -12.9 mol
Note: The negative sign indicates that we need a decrease in freezing point, rather than an increase. However, the number of moles cannot be negative, so we ignore the sign when calculating the quantity. Therefore, you would need approximately 12.9 moles of NaCl to lower the freezing point of 1 liter of water by 24 degrees Celsius.