Question

Assume that seawater is an aqueous solution at 3.50% NaCl. If the density of seawater is 1.025g /ml at 20 degrees Celcius, calculate your:

a) molality

b) molarity

c) parts per million (ppm)

Answer 1

a)

Let mass of solution be 1 Kg = 1000 g

mass of NaCl = 3.5 % of mass of solution

= 3.5*1000.0/100

= 35.0 g

mass of solvent = mass of solution - mass of solute

mass of solvent = 1000 g - 35.0 g

mass of solvent = 965.0 g

mass of solvent = 0.965 Kg

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 35.0 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(35 g)/(58.44 g/mol)

= 0.5989 mol

m(solvent)= 0.965 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.5989 mol)/(0.965 Kg)

= 0.6206 molal

Answer: 0.621 molal

b)

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 1.025 g/mL

use:

mass = density * volume

= 1.025 g/mL *1*10^3 mL

= 1.025*10^3 g

This is mass of solution

mass of NaCl = 3.5 % of mass of solution

= 3.5*1025.0/100

= 35.875 g

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 35.875 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(35.88 g)/(58.44 g/mol)

= 0.6139 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 0.6139/1

= 0.6139 M

Answer: 0.614 M

c)

Let mass of solution be 1 Kg = 1000 g

mass of NaCl = 3.5 % of mass of solution

= 3.5*1000.0/100

= 35.0 g

ppm = mass of NaCl * 10^6 / mass of solution

= 35.0*10^6 / 1000

= 35000 ppm

Answer: 35000 ppm