Question

A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m^3. Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration.

(a) 15.7%, 23.1%, 0.058, 3.88 moles in m^3

(b) 14.5%, 22.4%, 0.059, 2.88 moles in m^3

(c) 16.7%, 22.1%, 0.058, 3.77 moles in m^3

(d) 14.7%, 23.1%, 0.059, 2.77 moles in m^3

I had been asked this question by my college director while I was bunking the class.

Enquiry is from Law of Conservation of Mass in division Material Balance of Bioprocess Engineering

Answer 1

Right answer is (c) 16.7%, 22.1%, 0.058, 3.77 moles in m^3

To explain I would say: (i) Weight fraction:

20/(100+20) = 0.167; %weight/weight = 16.7%

(ii) Weight/volume:

A density of 1323kg/m^3 means that lm^3 of solution weighs 1323kg, but 1323kg of salt solution contains

(20×1323kg of salt) / (100+20) = 220.5 kg salt/m^3

1 m^3 solution contains 220.5 kg salt.

Weight/volume fraction = 220.5 / 1000 = 0.2205

And so weight / volume = 22.1%

(iii) Moles of water = 100 / 18 = 5.56

Moles of salt = 20 / 58.5 = 0.34

Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058

(iv) The molar concentration (M) is 220.5/58.5 = 3.77 moles in m^3.