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This is not a question kid
Answer:
Freezing T° of solution = - 4.52°C
Explanation:
ΔT = Kf . m . i
That's the formula for colligative property about freezing point depression.
Li₂O is an oxide that can not be dissociated but, if we see it's a ionic compound.
Li₂O → 2Li⁺ + O⁻²
3 moles of ions have been formed. Ions dissolved in solution are i, what we call Van't Hoff factor.
m is molality → 0.811 m, this is data
Kf →Cryoscopic constant, for water is 1.86 °C/m
and ΔT = Freezing T° of pure solvent - Freezing T° of solution
We replace: 0°C - Freezing T° of solution = 1.86°C/m . 0.811 m . 3
Freezing T° of solution = - 4.52°C
The molar mass should be 14.898g/mol.
I used the equation 100gx(1/xg/mol)x(1mol/2mol)x(16g/mol+x)/1)=103.7 and solve for x.
I found that equation using stoichiometry and the equation 2m+1/2O2-->m2O
The molar mass of the metal I set to x and the molar mass of the metal oxide is 16+x.
