If you purchase four 750 gram jars of table salt (NaCl) and dump them all onto 1,000 kg of snow, how much would the freezing point be lowered after all of the salt has dissolved?
mass of NaCl = 750 g mass of water = 1000 kg Kf = - 1.86 °C moles of NaCl = [tex] \frac{mass (g)}{Molar mass} [/tex] = [tex] \frac{750 g}{58.5 g/mol} [/tex] = 12.8 moles molality of NaCl = [tex] \frac{moles of solute (NaCl)}{mass of solvent (kg)} [/tex] = [tex] \frac{12.82}{1000} = 0.01282 mole/kg [/tex] Δ Tf = Kf * m = -1.86 * 0.01282 = - 0.024 °C
