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1-Common commercial acids and bases are aqueous solutions with the following pro

ID: 995528 • Letter: 1

Question

1-Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO3), density 1.42 g/cm3 and 68% solute by mass. Calculate the molality.

2-Common commercial acids and bases are aqueous solutions with the following properties: HCl, density 1.19 g/cm3 and 32% solute by mass. Calculate the mole fraction

3-Common commercial acids and bases are aqueous solutions with the following properties: nitric acid (HNO3), density 1.42 g/cm3 and 68% solute by mass. Calculate the mole fraction.

Explanation / Answer

1) the molality is given as

Molality = Moles of solute / Kg of solvent

Given :

Mass % = Mass of nitric acid / Mass of solution = 68 g / 100 g of solution

So in 100 g of solution there will be 68 g of nitric acid and rest water(32 g)

so 32 gram contains 68 g of nitric acid

1000 gram will have = 68 X 1000 / 32 g of nitric acid = 2125 grams

Moles of nitric acid = Mass / molecular weight = 2125 / 63 = 33.73

Molality = Moles of nitric acid / weight of solvent in Kg = 33.73 / 1 = 33.73 Molal

2) given :

Density = 1.19 g / cm^3

Mass % = 32%

To calculate : Mole fraction

mole fraction = Moles of solute / moles of solute + moles of solvent

Let us start with density , 1.19 g / cm^3

Mass of solute in 100 of solution = 32 grams

Moles = Mass / Molecule weight = 32 / 36.5 = 0.876

Mass of solvent in 100 g of solution = 68 g

Moles = 68 / 18 = 3.778

So molec fraction = 0.876 / 0.876 + 3.778 = 0.188

3) we can take help from above problem

in 100 g of solution there will be 68 g of nitric acid and rest water(32 g)

Moles of solute = Mass of solute/ Mol wt = 68 / 63 = 1.079

Moles of solvent = Mass of solvent / Mol wt = 32 /18 = 1.778

Mole fraction = Moles of solute / Moles of solute + moles of solvent = 1.079 / 2.857 = 0.378