Question

An aqueous solution contains 4.6 % NaCl by mass.

Part A
Calculate the molality of the solution

Part B
Calculate the mole fraction of the solution.

Answer 1

0.Let's assume we have 100 gram of this solution

amount of NaCl present in it will be 4.6 grams

moles of NaCl present =4.6/58.5 =0.079

amount of H₂O present in it = (100-4.6)gram = 95.4 grams

moles of H₂O present = 95.4/18=5.3

Part A:

Molality is number of moles in 1000 gram of solute = moles /mass of solute (in kg)

Molality = $\dfrac{0.079*1000}{95.4}=\dfrac{79}{95.4}=0.83$

Part B:

mole fraction = $\dfrac{moles \ of\ solute }{total \ moles }$$=\dfrac{0.079}{0.079+5.3}=0.079/5.379=$$0.015$