What is the new boiling point if 25 g of nacl is dissolved in 1.0 kg of water

The correct answer for the question that is being presented above is this one: 

Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m 
delta Tf = Kfm Kf H2O = 1.86 degrees C/m 

We need to know the formula for Molality.
molality = mol solute / kg solvent 

We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. 

25 g NaCl / 58.5 g/mol = 0.427 mol 

Then, use the formula for molality. 

molality = mol solute / kg solvent 
= 0.427 / 1 
= 0.427 m 

Use now the formula to get the boiling point.

delta Tb = Kbm 
= (0.52)(0.427) 
= 0.22C 

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