calculate the molecular weight of a substance which freezes forms a 7.00% by mass solution in water which freezes at -0.89°C.
See above problem with C6H6.
To calculate the molecular weight of a substance that forms a 7.00% by mass solution in water, we can use the colligative property of freezing point depression.
The freezing point depression (∆Tf) is the difference between the freezing point of the pure solvent and the freezing point of the solution. In this case, the freezing point depression is -0.89°C.
The formula for freezing point depression is given by:
∆Tf = Kf * m
Where:
∆Tf = freezing point depression
Kf = cryoscopic constant (a characteristic property of the solvent)
m = molality of the solution (moles of solute per kilogram of solvent)
Since we are dealing with a solution in water, the molality (m) can be calculated using the formula:
m = (mass of solute) / (molecular weight of solute * mass of solvent)
Given that the solution is 7.00% by mass, we can assume we have 100 grams of the solution. Therefore, the mass of the solute is 7 grams (7.00% of 100 grams).
Now, let's plug in the values we have into the formula and solve for the molecular weight of the solute.
1. Calculate the molality (m):
mass of solute = 7 grams
mass of solvent (water) = mass of solution - mass of solute = 100 grams - 7 grams = 93 grams
m = (7 grams) / (molecular weight of solute * 93 grams)
2. Calculate the freezing point depression (∆Tf):
∆Tf = -0.89°C
3. Determine the cryoscopic constant (Kf) for water. The cryoscopic constant for water is approximately 1.86 °C/m.
4. Rearrange the freezing point depression formula to solve for the molecular weight of the solute:
molecular weight of solute = (mass of solute) / (m * ∆Tf / Kf)
Now, substitute the known values:
molecular weight of solute = 7 grams / (m * -0.89°C / 1.86 °C/m)
Simplify the equation and calculate:
molecular weight of solute = (7 grams * 1.86 °C/m) / (m * -0.89°C)
= -13.02 grams * °C/m / m
Therefore, the molecular weight of the solute is -13.02 grams * °C/m.