Question

point depression data allow us to determine how many moles of particles are dissolved in solution. of 0500 grams of unknown and 20.0 grams of benzene freezes at 2.0O Trand K for benzene (see Table in lab procedure) to find the molality (mol/kg) of the solution, moles, and molar mass of the unknown substance? (Tt for Benzene is 5.5°C and Kf - 5.12C m) A 068 m, 36 6 mol, 0.014 g/mol 068 m 0.014 mol, 36.6 g/mol co 014m, 068 mol, 36.6 g/mol D None of these

plz answer as soon as possible

Answer 1

Δ Tf = 5.5 oC - 2.0 oC = 3.5 oC

use:

Δ Tf = Kf*mb

3.5 = 5.12 *mb

mb= 0.6836 molal

Answer: molality = 0.68 molal

m(solvent)= 20 g

= 2*10^-2 kg

use:

number of mol,

n = Molality * mass of solvent in Kg

= (0.6836 mol/Kg)*(2*10^-2 Kg)

= 1.367*10^-2 mol

Answer: number of mol = 1.4*10^-2 mol

mass(solute)= 0.500 g

use:

number of mol = mass / molar mass

1.367*10^-2 mol = (0.5 g)/molar mass

molar mass = 36.6 g/mol

Answer: molar mass = 36.57 g/mol

Answer:

B