Answer:
I have to weight 10,04 g of acetate sodium
Explanation:
This is the colligative propertie about elevation of boiling point
ΔT = Kb . m . i
ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure - we have this data 0,450°C
Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)
m means molality (moles of solute in 1kg of solvent)
i means theVan 't Hoff factor (degree of dissociation for a compound)
For the sodium acetate is 2
NaCH3COO ---> Na+ + CH3COO-
0,450°C = 0,52°C.kg/m . m . 2
0,450°C / (0,52°C.kg/m . 2) = m
0,432 m/kg = m
This number means I have 0,432 moles of acetate sodium, my solute in 1kg of water, my solvent. But I don't have 1000 g (1kg) I only have 283 g so let's make the rule of three:
1000 g _____ 0,432 moles
283 g ______ (283g .0,432m) / 1000g = 0,122 moles
Now that I have the moles of acetate sodium, I have to find the mass.
Moles . molar mass = mass
0,122 moles . 82.04 g/mol = 10,04 g
