Question

The density of an aqueous solution containing 15.0% ethanol (C2H5OH) by mass is 0.974 g/mL.

A) calculate the molality of this solution. B) calculate the solutions molarity.

C) what volume of the solution would contain 0.108 mole of ethanol?

Answer 1

A)

Let mass of solution be 1 Kg = 1000 g

mass of C2H5OH = 15.0 % of mass of solution

= 15.0*1000/100

= 150.0 g

mass of solvent = mass of solution - mass of solute

mass of solvent = 1000 g - 150.0 g

mass of solvent = 850.0 g

mass of solvent = 0.85 Kg

Molar mass of C2H5OH,

MM = 2*MM(C) + 6*MM(H) + 1*MM(O)

= 2*12.01 + 6*1.008 + 1*16.0

= 46.068 g/mol

mass(C2H5OH)= 150.0 g

use:

number of mol of C2H5OH,

n = mass of C2H5OH/molar mass of C2H5OH

=(1.5*10^2 g)/(46.07 g/mol)

= 3.256 mol

m(solvent)= 0.85 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(3.256 mol)/(0.85 Kg)

= 3.831 molal

Answer: 3.83 molal

B)

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 0.974 g/mL

use:

mass = density * volume

= 0.974 g/mL *1*10^3 mL

= 9.74*10^2 g

This is mass of solution

mass of C2H5OH = 15.0 % of mass of solution

= 15.0*974.0/100

= 146.1 g

Molar mass of C2H5OH,

MM = 2*MM(C) + 6*MM(H) + 1*MM(O)

= 2*12.01 + 6*1.008 + 1*16.0

= 46.068 g/mol

mass(C2H5OH)= 146.1 g

use:

number of mol of C2H5OH,

n = mass of C2H5OH/molar mass of C2H5OH

=(1.461*10^2 g)/(46.07 g/mol)

= 3.171 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 3.171/1

= 3.171 M

Answer: 3.17 M

C)

use:

M = number of mol / volume

3.171 = 0.108 / volume

volume = 0.0341 L

volume = 34.1 mL

Answer: 34.1 mL