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.

a) 132 g NaCl consumed at 1.5 Atm & 290K

b) 93.2 g NaCl consumed at STP

Explanation:

F₂ + 2NaCl => 2NaF + Cl₂

Given 18L F₂(g) at 1.5Atm & 290K

moles F₂(g) = PV/RT = (1.5Atm)(18L)/(0.08206L·Atm/mol·K)(290K) = 1.135 mol F₂(g) consumed

moles NaCl consumed = 2(moles F₂) = 2(1.135) mol NaCl = 2.269 mol NaCl

mass NaCl = 2.269 mol NaCl x 58 g/mol = 131.61 g ≅ 132 g consumed at 1.5 Atm & 290K

Given 18L F₂(g) - same volume as in part a -  at STP

moles F₂(g) = PV/RT = (1.0Atm)(18L)/(0.08206L·Atm/mol·K)(273K) = 0.803 mol F₂(g) consumed

moles NaCl consumed = 2(moles F₂) = 2(0.803) mole NaCl = 1.607 mol NaCl

mass NaCl = 1.607 mol NaCl x 58 g/mol = 93.2 g NaCl consumed at STP