Question

* Calculate the concentration of argon in the Pacific Ocean given that the partial pressure of argon is 0.00934 atm. The Henry’s law constant for argon in water at 25ºC is 1.4 x 10-3 mol/L·atm.

* The solubility of potassium chloride (KCl) is 34.2 g per 100.0 g of water at 20ºC.

Calculate the molality of a saturated solution of potassium chloride.

Is a 4.00 m solution of potassium chloride in water at 20ºC saturated, supersaturated, or unsaturated?

Answer: saturated, supersaturated, or unsaturated

Is a 6.00 m solution of potassium chloride in water at 20ºC saturated, supersaturated, or unsaturated?

Answer: saturated, supersaturated, or unsaturated

* A supermarket brand bleach contains 4.88 g of sodium hypochlorite (NaOCl) in 100.0 g of water. Calculate the molality, mass percent, ppm, and mole fraction of sodium hypochlorite in this solution.

Molality of sodium hypochlorite in water

Mass percent of sodium hypochlorite in water

ppm of sodium hypochlorite in water

Mole fraction of sodium hypochlorite in water

Answer 1

The concentration of Ar = henry's constant * partial pressure of Ar

= 1.4 x 10-3 mol/L·atm.* 0.00934 atm = 13.1 x10^-6 mols/L
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2)
For saturated KCl following numbers are given
mass of solute= 34.2 g
mass of solvent= 100g= 0.1 Kg
using these data and molar mass of KCl we can determine molality

molality = mass of KCl/molar mass of KCl*mass of solvent in Kg
= 34.2 g/0.1 g*74.55 g/mol = 4.588 m
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3)
4 molal solution means 4 moles in 1000 g of water

4mole KCl= 4*74.55 g/mol= 298.2 g kcl

So in 1000 g water there is 298.2 g KCl
Thus in 100 g water there will be 29.82 g KCl
given The solubility of KCl is 34.2 g per 100.0 g of water at 20ºC. This makes saturated
solution .

Since 29.82 g is smaller than 34.2 g :

The solution is unsaturated
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6 molal solution means 6 moles in 1000 g of water

6 mole KCl= 6*74.55 g/mol= 447.3 g kcl

So in 1000 g water there is 447.3 g KCl
Thus in 100 g water there will be 44.73 g KCl

So the solution is sper saturated
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5)

Mass of solute= 4.88 g
mass of water = 100 g= 0.1 Kg
molality = mass of NaOCl/molar mass of NaOCl*mass of solvent in Kg

= 4.88 g/74.44 g/mol*0.1 kg = 0.655 m
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use following formula to get mass %
Mass percent = mass of NaOCl/total mass of solution ]*100
= [4.88 g/100+4.88 ]*100 = 4.653 %
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concentration in ppm= grams of solute/grams of solution *1000000

concentration in ppm = 4.88g/100 g*1000000 = 48800 ppm

***************

to get mole fraction we need # moles of NaOCl and that of water

#of moles of NaOCl; Na = mass of NaOCl/molar mass =4.88g/74.44 g/mol==0.0655mols

# of moles of water; Nb = mass of water/molar mass of water
= 100 g/18 g/mol =5.55 mol

mole fraction Of NaOCl : Xa = Na/Na+Nb =0.0655mols/0.0655mols+ 5.55 mole
= 0.012
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All solved :))