Answer to Question #178779 in Chemistry for Karley Eilts

Question

What is the boiling point elevation of 45.6 g of NaCl dissolved in 650 g of water?

Answer 1

deltaT = K_bm

where:

       deltaT is the change in the boiling point of the solvent,

        K_b is the molal boiling point elevation constant (for water K_b=0.512^oC⋅m^-1)

        m is the molal concentration of the solute in the solution.

Cm = n/m

n = m/M

M (NaCl) = 58.44 g/mol

n (NaCl) = 45.6 / 58.44 = 0.8 mol

Cm (NaCl) = 0.8 / 0.65 = 1.2 m

deltaT = 0.512⋅1.2 = 0.6^oC