How many grams of pyrazine would have to be dissolved in 1.5kg of carbon tetrachloride to lower the freezing point by 4.4°C? Take kf of CCl4 to be 30°/molar
To determine the number of grams of pyrazine needed to lower the freezing point of carbon tetrachloride by a certain amount, we need to use the equation for freezing point depression:
ΔTf = Kf * m * i
where:
ΔTf is the freezing point depression (given as 4.4°C or 4.4 K in this case),
Kf is the freezing point constant of carbon tetrachloride (given as 30°C/molar),
m is the molality of the solution (mol solute per kg solvent),
and i is the van't Hoff factor, which represents the number of particles produced per formula unit (pyrazine is a non-electrolyte, so i = 1).
First, let's calculate the molality (m) of the solution:
m = n_solute / mass_solvent
Given that the mass of carbon tetrachloride is 1.5 kg and the molecular weight of carbon tetrachloride (CCl4) is 153.82 g/mol, we can calculate the moles of carbon tetrachloride:
n_solute = mass_solvent / molecular_weight_solvent
= 1500 g / 153.82 g/mol
≈ 9.75 mol
Next, we can plug the values into the freezing point depression equation to solve for the moles of pyrazine:
ΔTf = Kf * m * i
4.4 K = (30°C/molar) * (mol solute / 1.5 kg) * 1
Rearranging the equation to solve for mol solute:
mol solute = (ΔTf * 1.5 kg) / (30°C/molar)
= (4.4 K * 1.5 kg) / (30°C/molar)
≈ 0.22 mol
Finally, we can calculate the mass of pyrazine using the known formula weight of pyrazine, which is 80.10 g/mol:
mass_solute = mol solute * molecular_weight_solute
≈ 0.22 mol * 80.10 g/mol
≈ 17.62 g
Therefore, approximately 17.62 grams of pyrazine would need to be dissolved in 1.5 kg of carbon tetrachloride to lower the freezing point by 4.4°C.