Question

What is the molality of a 2.46 M aqueous solution of CH3OH? (Density = 0.976 g/mL)

Answer 1

Let volume be 1 L
volume , V = 1 L


use:
number of mol ,
n = Molarity * Volume
= 0.246*1
= 0.246 mol
volume , V = 1 L
= 1*10^3 mL


density, d = 0.976 g/mL
use:
mass = density * volume
= 0.976 g/mL *1*10^3 mL
= 976.0 g
This is mass of solution

Molar mass of CH3OH,
MM = 1*MM(C) + 4*MM(H) + 1*MM(O)
= 1*12.01 + 4*1.008 + 1*16.0
= 32.042 g/mol

use:
mass of CH3OH,
m = number of mol * molar mass
= 0.246 mol * 32.042 g/mol
= 7.882 g
This is mass of solute

mass of solvent = mass of solution - mass of solute
= 976 - 7.8823
= 968.1177 g
= 0.96812 Kg

m(solvent)= 0.96812 Kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.246 mol)/(0.96812 Kg)
= 0.2541 molal
Answer: 0.254 molal