A 3 mol/L of hydrochloric acid (HCI) solution has 1.013 of specific gravity. Calculate
a) mass concentration of solution in lb/ft? b) weight fraction and percent by weight of each component
c) mole fraction and percent by mole of each component
d) molality (Give density of water = 1 g/cm). =
Answers
A 3 mol/L of hydrochloric acid solution has 1.013 of specific gravity.
- (a) mass concentration of solution in lb/ft
- (b) weight fraction and percent by weight of each component
- (c) mole fraction and percent by mole of each component
- (d) molality
(a) here, 3 mol of hydrochloric acid is present in the 1 L of solution.
so, the mass of hydrochloric acid = no of moles of HCl × molar mass of HCl
= 3 × 36.5 = 109.5 g
now the volume of solution = 1L = 1000 cm³
so the mass concentration of solution in g/cm³ = mass of solute in g/volume of solution in cm³
= 109.5/1000 g/cm³ = 0.1095 g/cm³
∵ 1lb/ft³ = 0.0160185 g/cm³
so, 0.1095 g/cm³ = 0.0160185/0.1095 = 0.146287 ≈ 0.15 lb/ft³
Therefore the mass concentration of solution in lb/ft³ is 0.15 lb/ft³.
(b) weight of HCl = 109.5 g
weight of solution = density of solution × volume of solution
= specific density × density of water × volume of solution
= 1.013 × 1 g/cm³ × 1000 cm³
= 1013 g
so the weight fraction of HCl = 109.5/1013 ≈ 0.1081
∴ percentage by weight of HCl = weight fraction × 100 = 10.81 %
weight fraction of solvent = 1 - weight fraction of solute
= 1 - 0.1081 = 0.8919
∴ percent by weight of solvent = 0.8919 × 100 = 89.19%
(c) no of moles of HCl = 3
mass of solvent = 1013 - 109.5 = 903.5 g
no of moles of solvent ( water) = 903.5/18 = 50.2
∴ mole fraction of HCl = 3/(3 + 50.2) = 3/53.2 = 0.05639
∴ percent by mole of HCl = 0.05639 × 100 = 5.639 %
mole fraction of solvent = 1 - mole fraction of solute = 1 - 0.05639 = 0.94361
percent by mole of solvent = 0.94361 × 100 = 94.361%
(d) molality = no fo moles of solute/mass of solvent in Kg
= ( 3mol)/(903.5/1000 kg )
= 3000/903.5 = 3.32 molal