Question

At what temperature will water freeze on a standard, single car driveway with 0.50 cm thick ice (assume 105 L of water) that has been salted with 11.0 kg CaCl2? *AȚf K.cm i, K,--1.86 Cim, assume d = 1 g/mL

Answer 1

Water will freeze at a temperature of -5.27^oC

Explanation

Volume of water = 105 L

Mass of water = (volume of water) * (density of water)

Mass of water = (105 L) * (1 kg/L)

Mass of water = 105 kg

mass CaCl₂ = 11.0 kg = 11000 g

moles CaCl₂ = (mass CaCl₂) / (molar mass CaCl₂)

moles CaCl₂ = (11000 g) / (111 g/mol)

moles CaCl₂ = 99.1 mol

molality CaCl₂ = moles CaCl₂ / mass of water (kg)

molality CaCl₂ = 99.1 mol / 105 kg

molality CaCl₂ = 0.944 m

According to equation for depression in freezing point,

T_f = = i * K_f * m

Water will freeze at a temperature of -5.27 degree C Explanation Volume of water = 105 L Mass of water = (volume of water) * (density of water) Mass of water = (105 L) * (1 kg/L) Mass of water = 105 kg mass CaCl_2 = 11.0 kg = 11000 g moles CaCl_2 = (mass CaCl_2)/(molar mass CaCl_2) moles CaCl_2 = (11000 g)/(111 g/mol) moles CaCl_2 = 99.1 mol molality CaCl_2 = moles CaCl_2/mass of water (kg) molality CaCl_2 = 99.1 mol/105 kg molality CaCl_2 = 0.944 m According to equation for depression in freezing point, delta T_f = = i * K_f * m where delta T_f = decrease in freezing point i = Van't Hoff factor = number of ions after dissociation = 3 (for CaCl_2) K_f = -1.86 degree C/m m = molality of solute = 0.944 m Substituting the values delta T_f = (3) * (- 1.86 degree C/m)

where T_f = decrease in freezing point

Water will freeze at a temperature of -5.27 degree C Explanation Volume of water = 105 L Mass of water = (volume of water) * (density of water) Mass of water = (105 L) * (1 kg/L) Mass of water = 105 kg mass CaCl_2 = 11.0 kg = 11000 g moles CaCl_2 = (mass CaCl_2)/(molar mass CaCl_2) moles CaCl_2 = (11000 g)/(111 g/mol) moles CaCl_2 = 99.1 mol molality CaCl_2 = moles CaCl_2/mass of water (kg) molality CaCl_2 = 99.1 mol/105 kg molality CaCl_2 = 0.944 m According to equation for depression in freezing point, delta T_f = = i * K_f * m where delta T_f = decrease in freezing point i = Van't Hoff factor = number of ions after dissociation = 3 (for CaCl_2) K_f = -1.86 degree C/m m = molality of solute = 0.944 m Substituting the values delta T_f = (3) * (- 1.86 degree C/m)

i = Van't Hoff factor = number of ions after dissociation = 3 (for CaCl₂)

K_f = -1.86 ^oC/m

m = molality of solute = 0.944 m

Substituting the values

T_f = (3) * (-1.86 ^oC/m) * (0.944 m)

Water will freeze at a temperature of -5.27 degree C Explanation Volume of water = 105 L Mass of water = (volume of water) * (density of water) Mass of water = (105 L) * (1 kg/L) Mass of water = 105 kg mass CaCl_2 = 11.0 kg = 11000 g moles CaCl_2 = (mass CaCl_2)/(molar mass CaCl_2) moles CaCl_2 = (11000 g)/(111 g/mol) moles CaCl_2 = 99.1 mol molality CaCl_2 = moles CaCl_2/mass of water (kg) molality CaCl_2 = 99.1 mol/105 kg molality CaCl_2 = 0.944 m According to equation for depression in freezing point, delta T_f = = i * K_f * m where delta T_f = decrease in freezing point i = Van't Hoff factor = number of ions after dissociation = 3 (for CaCl_2) K_f = -1.86 degree C/m m = molality of solute = 0.944 m Substituting the values delta T_f = (3) * (- 1.86 degree C/m)

T_f = -5.27 ^oC

Water will freeze at a temperature of -5.27 degree C Explanation Volume of water = 105 L Mass of water = (volume of water) * (density of water) Mass of water = (105 L) * (1 kg/L) Mass of water = 105 kg mass CaCl_2 = 11.0 kg = 11000 g moles CaCl_2 = (mass CaCl_2)/(molar mass CaCl_2) moles CaCl_2 = (11000 g)/(111 g/mol) moles CaCl_2 = 99.1 mol molality CaCl_2 = moles CaCl_2/mass of water (kg) molality CaCl_2 = 99.1 mol/105 kg molality CaCl_2 = 0.944 m According to equation for depression in freezing point, delta T_f = = i * K_f * m where delta T_f = decrease in freezing point i = Van't Hoff factor = number of ions after dissociation = 3 (for CaCl_2) K_f = -1.86 degree C/m m = molality of solute = 0.944 m Substituting the values delta T_f = (3) * (- 1.86 degree C/m)

Water will freeze at a temperature of -5.27^oC