The translation of the question is
Commercial solutions of HCl (MP = 36.5 g / mol) are typically 39.0% W / W of this acid in water. Determine the molality of the HCl solution if its density is equal to 1.20 g / ml.
Answer:
Molality is 17.51 molal
Explanation:
We are given 39% w/w of HCl, this means for every 100g of this solution we have 39g of HCl in it.
to ease the calculation we assume we have 1L(1000ml) of this solution is present.
Density = mass /volume
1.20 = mass/1000
mass = 1200 g of solution
and we are given that 39% of this solution is will contain HCl
so
1200 * 39/100 = 468g of HCl present
no of moles of HCl = 468/36.5 = 12.82moles
mass of solution = mass of water + mass of acid
mass of water = mass of solution - mass of acid
Mass of water = 1200 - 468 = 732g of water
Molality = moles /masss of water in kg
= 12.82 *1000/732 =17.51molal
