Question

2 attempts left Check my work Enter your answer in the provided box. A solution of 1.15 g of solute dissolved in 25.0 mL of H20 at 25°C has a boiling point of 100.750°C. What is the molar mass of the solute if it is a nonvolatile nonelectrolyte and the solution behaves ideally (d of H20 at 25°C = 0.997 g/mL)? g/mol

Answer 1

Normal boiling point of water is 100 oC
So,
Δ Tb = 110.750 - 100 = 0.750 oC
use:
Δ Tb = Kb*mb
0.75 = 0.512 *mb
mb= 1.4648 molal


m(solvent)= density * volume
= 0.997 g/mL * 25.0 mL
= 24.925 g
= 2.4925*10^-2 kg

use:
number of mol,
n = Molality * mass of solvent in Kg
= (1.465 mol/Kg)*(2.4925*10^-2 Kg)
= 3.651*10^-2 mol

mass(solute)= 1.15 g


use:
number of mol = mass / molar mass
3.651*10^-2 mol = (1.15 g)/molar mass
molar mass = 31.5 g/mol
Answer: 31.5 g/mol