the freezing point of benzene C6H6 is 5.5000C at 1 atmosphere.
In a laboratory experiment, students synthesized a new compound and found that when 10.70 grams of the compound were dissolved in 294.0 grams of benzene, the solution began to freeze at 4.9740C. The compound was also found to be nonvolatile and a non-electrolyte.
what is the molecular weight?
aftr you find the m of the unknown then i don't know how to get the mol of the unknown?
m = mols/kg solvent. You know m and kg solvent, solve for mols.
mols = grams/molar mass. You know mols and grams, solve for molar mass.
To find the molecular weight of the compound, we can use colligative properties, specifically the freezing point depression equation:
ΔT = K * m * i
Where:
ΔT = change in freezing point
K = constant (specific to the solvent)
m = molality (moles of solute/kg of solvent)
i = van't Hoff factor (number of particles into which a solute dissociates)
In this case, since the compound is nonvolatile and a non-electrolyte, i will be equal to 1.
First, we need to calculate the molality (m) of the compound in the benzene solution by dividing the moles of solute by the kilograms of solvent:
Moles of solute = mass of compound / molar mass of compound
From the given information, the mass of the compound is 10.70 grams. We need to convert this to moles using the molar mass of the compound.
Now, to calculate the molar mass of the compound, we can rearrange the freezing point depression equation:
molar mass of compound = mass of compound / (moles of solute / molality)
Next, we convert the given mass of benzene into kg:
mass of benzene = 294.0 grams = 0.294 kg
Now, we substitute the values into the freezing point depression equation:
ΔT = K * m
Solving for K:
K = ΔT / m
From the given information, the change in freezing point (ΔT) is 5.5000C - 4.9740C = 0.5260C.
Next, substitute the values into the equation for K:
K = 0.5260C / m
Now, we substitute the values into the molar mass equation:
molar mass of compound = 10.70 g / (moles of solute / (0.294 kg * 0.5260C))
By rearranging the equation, we can solve for moles of solute:
moles of solute = (10.70 g * (0.294 kg * 0.5260C)) / molar mass of compound
Finally, substitute this value into the equation for molar mass of the compound:
molar mass of compound = 10.70 g / [(10.70 g * (0.294 kg * 0.5260C)) / molar mass of compound]
By rearranging and simplifying the equation, we can solve for the molar mass of the compound.