Answer to Question #165170 in General Chemistry for Eli

Question

What is the boiling point of the solution formed by dissolving .25 mole of a nonvolatile, nonelectrolyte in 250g of water at 1 atm

Answer 1

$$\Delta\ T b=Tb(solution)-Tb(pure solvent)$$

$$\Delta\ T b=E*Cm$$

E - ebullioscopic constant (0.52 for water)

$$Cm=m(solute)*1000/(M(solute)*m(solvent))=n*1000/m(solvent)$$

$$\Delta Tb=0.52*0.25*1000/250=0.52°C$$

$$Tb(solution)=100+0.52=100.52 °C$$