Question

PEIT A What mass of salt (NaCl) should you add to 1.90 L of water in an ice cream maker to make a solution that freezes at-11.4 C ? Assume complete dissociation of the NaCl and density of 1.00 g/mL for water. VAE ? m(NaCl) Request Answer Submit

Answer 1

m(NaCl) = 340 g

Explanation

freezing point depression = normal freezing point - depressed freezing point

freezing point depression = 0.0 ^oC - (-11.4 ^oC)

freezing point depression = 11.4 ^oC

molality of NaCl = (freezing point depression) / [(i) * (K_f)]

molality of NaCl = (11.4 ^oC) / [(2) * (1.86 ^oC/m)]

molality of NaCl = 3.06 m

moles NaCl = (molality of NaCl) * (mass of water in kg)

moles NaCl = (3.06 m) * (1.90 kg)

moles NaCl = 5.82 mol

mass NaCl = (moles NaCl) * (molar mass NaCl)

mass NaCl = (5.82 mol) * (58.44 g/mol)

mass NaCl = 340.27 g