A solution is made by dissolving 1.00 moles of sodium chloride in 155 grams of water. If the molal boiling point constant water is .51 degrees C/m, what would be the boiling point of this solution?

Question

anonymous · Answer

$$DeltaTb=Kb*m$$
m=moles/kg of solvent
$$m=1/.155$$=6.452
$$DeltaTb=6.452*0.51=3.291$$
$$DeltaTb=Tbfinal-Tboriginal$$
$$Tbfinal=DeltaTb+Tboriginal=3.29+100$$
103.29 degrees Celsius is the new boiling point