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

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.