Question

A solution of 1.80 g of solute dissolved in 25.0 mL of H2O at 25 C has a boiling point of 100.50 C. What is the molar mass of the solute if it is a nonvolatile nonelectrolyte and the solution behaves ideally (d of H2O at 25 C = 0.997 r,mL)?

Answer 1

we have below equation to be used:

delta Tb = Kb*mb

0.5 = 0.512 *mb

mb= 0.9766 molal

mass of solvent = density * volume

= 0.997 g/mL*25.0 mL

= 249.25 g

= 0.2492 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

number of mol,

n = Molality * mass of solvent in Kg

= (0.9766 mol/Kg)*(0.2492 Kg)

= 0.2434 mol

mass of solute = 1.80 g

we have below equation to be used:

number of mol = mass / molar mass

0.2434 mol = (1.8 g)/molar mass

molar mass = 7.395 g/mol

Answer: 7.395 g/mol