After the snow storm, salttrucks are usually sent to spread salt, usually sodium

Question

After the snow storm, salttrucks are usually sent to spread salt, usually sodium chloride, onroads and
sidewalks to prevent ice from forming in the sub-zero weather. Thefreezing point of water is exactly 0ºC but the melting pointcan be lowered by the addition of sodium chloride.
(a) What would be the freezing point of the solution if 175 g ofsodium chloride (NaCl) is added to 1.00 kg of water? Kf of water is1.86 ºC·kg/mol.(b) Calcium chloride (CaCl2) is also used for de-icing. Whatwould be the freezing point of the solution if 175 g of calciumchloride (CaCl2) is added to 1.00 kg of water?

Explanation / Answer

Formula:                   Tf = Kf * m                   Tf = Tpure - Tsoln                   Tpure for water = 0 0C a ) Data:                Mass of NaCl = 175 g           Molarmass of NaCl = 58.442 g / mol Number of moles of NaCl = 175 g / 58.442 g /mol                                           = 2.994 mols                                  m     = 2.994 mols / 1.00 kg                                          = 2.994 mol / kg                                   Kf   =  1.86 ºC·kg/mol                               Tf = 1.86 ºC·kg/mol x 2.994mol / kg                                           =5.56 ºC                                Tsoln = -5.56 ºC b) Data:                Mass of CaCl2 = 175 g           Molarmass of CaCl2 = 110.986 g / mol Number of moles of CaCl2 = 175 g/ 110.986 g /mol                                           =  1.576mols                                  m     = 1.576 mols / 1.00 kg                                          = 1.576 mol / kg                                   Kf   =  1.86 ºC·kg/mol                               Tf = 1.86 ºC·kg/molx 1.576 mol / kg                                           = 2.93ºC                                Tsoln = -2.93 ºC                Mass of CaCl2 = 175 g           Molarmass of CaCl2 = 110.986 g / mol Number of moles of CaCl2 = 175 g/ 110.986 g /mol                                           =  1.576mols                                  m     = 1.576 mols / 1.00 kg                                          = 1.576 mol / kg                                   Kf   =  1.86 ºC·kg/mol                               Tf = 1.86 ºC·kg/molx 1.576 mol / kg                                           = 2.93ºC                                Tsoln = -2.93 ºC                        

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