Question

Common commercial acids and bases are aqueous solutions with the following properties:

	Density in g/cm3	Mass percent of solute
Hydrochloric acid	1.19	38
Nitric acid	1.42	70
Sulfuric acid	1.84	95
Acetic acid	1.05	99
Ammonia	0.90	28

Calculate the molarity, molality, and mole fraction of

each of these reagents.

Short answer

Molarity: a) Hydrochloric acid = 9.72M

b) Nitric acid = 14.88M

c) Sulfuric acid = 17.12M

d) Acetic acid = 16.62M

e) Ammonia = 8.75M

Molality : a) Hydrochloric acid = 16.80m

b) Nitric acid = 64m

c ) sulfuric acid = 193.8m

d) Acetic acid = 1583m

e) Ammonia = 12.41m

Mole fraction : a) Hydrochloric acid = 0.232

b) Nitric acid = 0.535

c) Sulfuric acid = 0.777

d) Acetic acid = 0.996

e) Ammonia = 0.184

Step-by-step solution

 Hydrochloric acid :

Molarity of a reagent =$\frac{\text{moles of solute}}{\text{volume of solution in litres}}$

We assume mass of solution to be 1 kg or 1000g

Mass of hydrochloric acid= 380g

Number of moles of hydrochloric acid=$\frac{380}{36.45}$= 10.42mol

Mass of water = 1000-380 = 620g

Number of moles of water =$\frac{620}{18}$= 34.44mol

Density of water =1$g/c{m}^3$

Volume of water = 620mL

Volume of hydrochloric acid= density of hydrochloric acid mass of hydrochloric acid

= 1.19 380 = 452.2

Volume of solution in L=$\frac{620+452.2}{1000}$ = 1.072mL

Molarity= $\frac{10.42}{1.072}$= 9.72M

$\text{molality =}\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Molality =$\frac{10.42}{0.62}$= 16.80m

${\text{mole fraction pf a componentx}}_a=\frac{n_a}{(n_a+n_b)}$

Moles of hydrochloric acid = 10.42mol

Moles of water = 34.44mol

Mole fraction= $\frac{10.42}{10.42+34.44}$= 0.232

             Nitric acid

Molarity of a reagent =$\frac{\text{moles of solute}}{\text{volume of solution in litres}}$

We assume mass of solution to be 1 kg or 1000g

Mass of nitric acid= 700g

Number of moles of nitric acid=$\frac{700}{36.45}$= 19.20mol

Mass of water = 1000-700 = 300g

Number of moles of water =$\frac{300}{18}$= 16.66mol

Density of water =1$g/c{m}^3$

Volume of water = 300 mL

Volume of nitric acid= density of nitric acid $\times$mass of nitric acid

= 1.42$\times$700=994

Volume of solution in L=$\frac{300+994}{1000}$ = 1.294L

Molarity=$\frac{19.20}{1.294}$ = 14.88M

$\text{molality =}\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Molality =$\frac{19.20}{0.3}$= 64m

${\text{mole fraction pf a componentx}}_a=\frac{n_a}{(n_a+n_b)}$

Moles of hydrochloric acid = 19.20mol

Moles of water = 16.66mol

Mole fraction=$\frac{19.20}{19.20+16.66}$= 0.535

Similarly,

Sulfuric acid

Molarity=17.12M

Molality=193.8m

Mole fraction=0.77

Acetic acid

Molarity=16.62M

Molality=1583m

Mole fraction=0.996

Ammonia

Molarity=8.75M

Molality=12.41m

Mole fraction =0.184