Question

An RV antifreeze is an aqueous solution containing 5.68 M proplyene glycol (MM=76.09 g/mol). If the density of the antifreeze is 1.080 g/mL, what is the molality of the solution?

Answer 1

proplyene glycol is C3H8O2

Let volume be 1 L
volume , V = 1 L


use:
number of mol ,
n = Molarity * Volume
= 5.68*1
= 5.68 mol
volume , V = 1 L
= 1*10^3 mL


density, d = 1.08 g/mL
use:
mass = density * volume
= 1.08 g/mL *1*10^3 mL
= 1080.0 g
This is mass of solution

Molar mass of C3H8O2 = 76.09 g/mol

use:
mass of C3H8O2,
m = number of mol * molar mass
= 5.68 mol * 76.09 g/mol
= 4.322*10^2 g
This is mass of solute

mass of solvent = mass of solution - mass of solute
= 1080 - 432.2139
= 647.7861 g
= 0.64779 Kg

m(solvent)= 0.64779 Kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(5.68 mol)/(0.64779 Kg)
= 8.768 molal
Answer: 8.77 molal