What are the new boiling point, and freezing point if: 25 g of NaCl is dissolved in 1.0 Kg of water?

justaguide | Certified Educator

When salt is dissolved in water the freezing point of water drops below 0 degree Celsius which is the freezing point of pure water and the boiling point rises above 100 degrees Celsius.

The exact change in the freezing point is given by dTf = Kf*cm and the change in the boiling point is given by dTb = Kb*cm, where Kf and Kb are constants for a liquid and cm is the the molality of salt.

Here, we have 25 g of salt being dissolved in a kilogram of water. The molecular mass of NaCl is 58.443 g/ mole. 25 g of salt is 0.4277 moles. The molality of salt in water is 0.4277.

The constant Kf for water is 1.858 and Kb is 0.521.

The change in the boiling point is 0.4277*0.521 = 0.222 degrees Celsius or the new boiling temperature is 100.22 degrees Celsius.

The change in the freezing point is 0.4277*1.858 = 0.7947 degrees Celsius or the new freezing point is -0.7947 degrees Celsius.