Compute the freezing point of this mixture (in °C).?
Homemade ice-cream is frozen by churning it in a bucket suspended in an ice-water-salt mixture. A typical mix calls for 2.50 kg of salt (NaCl) and 6.25 kg of ice.
Use the formula for freezing point depression: ΔTf = Kf * mB where Kf is the constant for water (1.86 (kg*K)/mol) and mB is the molality of the solute (mol/kg).
NaCl has a molar mass of 58.44 g/mol or 0.05844 kg/mol.
2.50 kg of NaCl therefore contains 44.779 moles of NaCl.
However, remember NaCl is dissociated in water, making Na+ and Cl- ions. Therefore, there are 44.779*2 = 89.56 moles of solute in your 6.25 kg of solvent. So the molality (moles of solute per kg of solvent) is 89.56/6.25 = 14.33 mB.
14.33*1.86 = 26.65 K.
The freezing point thus depressed by 26.65 K = 26.65 deg C. 0 - 26.65 = -26.65 deg C.
Answers & Comments
Verified answer
Use the formula for freezing point depression: ΔTf = Kf * mB where Kf is the constant for water (1.86 (kg*K)/mol) and mB is the molality of the solute (mol/kg).
NaCl has a molar mass of 58.44 g/mol or 0.05844 kg/mol.
2.50 kg of NaCl therefore contains 44.779 moles of NaCl.
However, remember NaCl is dissociated in water, making Na+ and Cl- ions. Therefore, there are 44.779*2 = 89.56 moles of solute in your 6.25 kg of solvent. So the molality (moles of solute per kg of solvent) is 89.56/6.25 = 14.33 mB.
14.33*1.86 = 26.65 K.
The freezing point thus depressed by 26.65 K = 26.65 deg C. 0 - 26.65 = -26.65 deg C.
im sorry idk,
but PLEASE help meee!!!
http://ca.answers.yahoo.com/question/index?qid=200...