What weight of glycerol C3H5(OH)3 molar mass is 92 would have been add to 1000g of water in order to lower the freezing point
To calculate the weight of glycerol that needs to be added to lower the freezing point of water, we need to use the equation for freezing point depression:
ΔT = K_f * m
Where:
ΔT is the change in freezing point
K_f is the cryoscopic constant (1.86 °C/m for water)
m is the molality of the solute (moles of solute per kg of solvent)
First, we need to calculate the molality of the solution by converting the given weight of water (1000g) to kilograms:
mass of water (kg) = 1000g / 1000 = 1kg
Next, we need to calculate the number of moles of glycerol using its molar mass. The molar mass of glycerol (C3H5(OH)3) is:
molar mass of glycerol = (3 * atomic mass of carbon) + (8 * atomic mass of hydrogen) + (3 * atomic mass of oxygen)
= (3 * 12.01 g/mol) + (8 * 1.01 g/mol) + (3 * 16.00 g/mol)
≈ 92.09 g/mol
Now, we can calculate the number of moles of glycerol:
moles of glycerol = mass of glycerol / molar mass of glycerol
Next, we need to calculate the change in freezing point (ΔT). Since the freezing point depression equation assumes an ideal solution, the change in freezing point (ΔT) is proportional to the molality (m). Therefore:
ΔT = K_f * m
Now, we substitute the known values into the equation:
ΔT = (1.86 °C/m) * m
To find the molality (m) of the solution, rearrange the equation:
m = ΔT / K_f
Here, we need to know the magnitude of the freezing point depression (ΔT). You haven't provided that information, so we cannot proceed further with the calculation. If you provide the value of ΔT, we can continue with the calculation and determine the weight of glycerol needed to achieve that freezing point depression.