Answer:
The molar mass for the solute is 180.8 g/mol
Explanation:
We apply the colligative property of freezing point depression.
The formula is T° freezing pure solvent - T° freezing solution = Kf . m
T° freezing pure solvent - T° freezing solution → ΔT
Kf = 1.86 °C/m
m is molality. We must determine with data given.
Molality are the moles of solute contained in 1 kg of solvent
0° - (-1.8°C) = 1.86 °C/m . m
1.8°C / 1.86 m/°C = m → 0.968 mol/kg
We multiply the molality by kg of solute, to determine the moles
100 g . 1kg/1000 g = 0.1kg → 0.968 mol/kg . 0.1 kg = 0.0968 moles
To find the molar mass we make g/ mol → 17.5 g / 0.0968 mol =
180.8 g/mol
Answer:
The molar mass of the nonelektrolyte is 180.8 g/mol
Explanation:
Step 1: Data given
Mass of non-elektrolyte = 17.5 grams
MAss of water = 100 grams = 0.1 kg
Freezing point of solution = -1.8 °C
Kf water = 1.86 °C/m
Step 2: Calculate molality
ΔT = i*kf*m
⇒ with ΔT = the freezing point depression = 1.8 °C
⇒ with i = the van't Hoff factor = 1
⇒ with kf = the freezing point depression constant = 1.86 °C/m
⇒ with m = molality
1.8 = 1* 1.86 * m
m =1.8 / 1.86
m = 0.9677 molal
Step 3: Calculate moles nonelectrolyte
molality = moles nonelectrolyte / mass water
0.9677 molal = moles / 0.1 kg
moles = 0.9677 molal * 0.1 kg
moles = 0.09677 moles
Step 4: calculate molar mass of the nonelectrolyte
Molar mass = mass / moles
Molar mass = 17.5 grams / 0.09677 moles
Molar mass = 180.8 g/mol
The molar mass of the nonelektrolyte is 180.8 g/mol
