You are asked to prepare an aqueous solution of ethylene glycol (HOCH2CH2OH) with a mole fraction of 0.111.?

Question

What is the molality of the resulting solution?

Answer 1

Atomic weights: C=12 H=1 O=16 C2H6O2=62 H2O=18

Let ethylene glycol be called EG

If the mole fraction of EG is 0.111, then that of water is 1.000 -0.111 = 0.889.

0.889molH2O x 18gH2O/1molH2O = 16gH2O

0.111molEG/16gH2O x 1000gH2O/1kgH2O = 6.93 mol EG/1kgH2O

This is the molality, because molality is number of moles solute (EG) in 1 kg solvent (H2O).

Answer 2

Let m be molality.

m = (weight of solute x 1000) / ( molar mass of solute x weight of solvent)

All weights are in g and molar mass is in g/mol.

Let X be the mole-fraction of ethylene glycol.
a is solvent and b is solute (ethylene glycol).
X = 0.111 = Nb/ Na + Nb
=> 111/1000 = Nb/Na +Nb
=> 111Na + 111Nb = 1000Nb
=> 111Na = 889Nb
=> Nb/Na = 111/889
=> (wt. of b/ molar mass of b) / ( wt. of a / molar mass of a) =111/889

Now, molar mass of a, H2O is 18g/mol
molar mass of b, HOCH2CH2OH is 62 g/mol
=> (wt. of b/62) / (wt.of a / 18) =111/889
=> wt. of b/ wt.of a = 0.43007

Using the above ratio in the general eqn of molality;

m = ( wt. of b) x 1000 / (wt.of a ) x molar mass of b
m = (0.43007 x1000)/ 62
=430.07/62
= 6.936 mol (Ans.)

Answer 3

Remove Your Moles, Warts & Skin Tags in 3 Days
Easily, Naturally And Without Surgery

Answer 4

Find out the moles.