Question

The percentage purity of commercial HCl is 35% (w/w) & the specific gravity of the solution is 1.18 g/mL. What is the molarity, molality and normality?

Answer 1

Hint: To solve the given question we need to find the volume of the solution from the specific gravity. The mass of the solute i.e. HCl will be 35g dissolved in 100 g of solution. The normality and Molarity of HCl will be equal as HCl is monoprotic.

Complete answer:
The %w/w of the given HCl solution is 35% which means that 35g of HCl is dissolved in 100g of solution. The specific gravity of the given solution is 1.18 g/mL. The volume of the solution can be given by the formula:
\[Volume = \dfrac{{Mas{s_{solution}}}}{{Sp.Gravity}} = \dfrac{{100}}{{1.18}} = 84.74mL\]
The no. of moles of the solute can be given by the formula: $Moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}Mass(g/mol)}}$
No. of moles of HCl dissolved (given that the molar mass of HCl = 36.5g/mol) $ = \dfrac{{35}}{{36.5}} = 0.9589mol$
Molarity of the solution is given as: $Molarity = \dfrac{{Mole{s_{solute}}}}{{Volum{e_{solution}}(L)}}$
The molarity of the given solution of HCl is $ = \dfrac{{0.9589 \times 1000}}{{84.74}} = 11.315M$
The molality of the solution can be given by the formula: $Molality = \dfrac{{Mole{s_{solute}}}}{{Mas{s_{solvent}}(kg)}}$
The mass of the solution is 100 g with 35 g of solute dissolved. The mass of the solvent will be $ = 100 - 35 = 65g$
Therefore, the molality will be $ = \dfrac{{0.9589 \times 1000}}{{65}} = 14.752m$
The normality of the HCl will be the same as the Molarity as HCl is a monoprotic acid. Therefore, the Normality of the HCl solution also be $ = 11.315N$

Note:
%w/w refers to the unit of solute (in grams) dissolved in 100 g of solution. %w/v refers to the grams of solute dissolved in 100mL of solution. For finding the molality the mass of the solvent has to be found in kg. Remember to convert the mass of solvent from g to kg.