How many grams of (nh4)3po4 need to be added to 500. g of h2o so that the freezing point of the solution is lowered to –8.3o c?

Answer is: mass of (NH₄)₃PO₄) is 333,76 g.
m(H₂O) = 500 g = 0,5 kg.
ΔT = 8,3°C = 8,3 K.
Kf(H₂O) = 1,853 K·kg/mol.
m((NH₄)₃PO₄) = ?
M((NH₄)₃PO₄) = 149 g/mol.
b = n ÷ m(H₂O).
ΔT = Kf · b.
b = 8,3 K ÷ 1,853 K·kg/mol.
b = 4,48 mol/kg.
n((NH₄)₃PO₄) = b · m(H₂O).
n((NH₄)₃PO₄) =4,48 mol/kg · 0,5 kg.
n((NH₄)₃PO₄) = 2,24 mol.
m((NH₄)₃PO₄) = 2,24 mol · 149 g/mol.
m((NH₄)₃PO₄) = 333,76 g.

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