Answer:
548 g/mol
Explanation:
The freezing point depression of a solvent occurs when a nonvolatile solute is added to it. Because of the interactions between solute-solvent, it is more difficult to break the bonds, so the phase change will need more energy, and the freezing point will drop, which is called cryoscopy.
The drop in temperature can be calculated by:
ΔT = Kf*W*i
Where Kf is the cryoscopy constant of the solvent, W is the molality, and i is the van't Hoff factor, which indicates the fraction of the solute that dissolves.
The molality represents how much moles (n) of the solute is presented in each kg of the solvent (m2), thus
W = n/m2
The number of moles is the mass of the solute (m1) in g, divided by the molar mass (M1) of it:
W = m1/(M1*m2)
So, by the data:
0.2214 = 0.632/(M1*0.00521)
0.00115M1 = 0.632
M1 = 548 g/mol
Answer:
Molecular weight of the solute = 0.5479 kg/mol or 547.9g/mol
Explanation:
Molarity,M is defined as the number of moles of solute dissolved in 1 liter of solution.
M = n/V
Where n is the number of moles
V is the volume of solution in liter
Molality,m is the number of moles of solute per kilogram of solvent.
molality = n/Ms
Where n is the number of solute
Ms is the kilogram of solvent in kg
In the information given,
Mass of solute = 0.632g
Molality, m = 0.2214
Mass of solvent, Ms = 0.00521kg
Using,
Molality, m = number of moles of solute/kilogram of solvent
Number of moles of solute = mass of solute in kg/molecular weight, M.W of solute in kg/mol
n = (0.632/1000)/M.W
Inputting all these values,
0.2214 = (6.32 x 10^-4/M.W)/0.00521
M.W = 6.32 x 10^-4/(0.00521 * 0.2214)
= 0.5479 kg/mol
= 547.9g/mol
