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
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