Answer:
Molality = 8.57 m
Explanation:
Given data:
Molarity of solution = 5.73 M
density = 0.9327 g/mL
Molality of solution = ?
Solution:
Molality = moles of solute / kg of solvent.
Kg of solvent:
Mass of 1 L solution = density× volume
Mass of 1 L solution = 0.9327 g/mL × 1000 mL
Mass of 1 L solution = 932.7 g
Mass of solute:
Mass of 1 L = number of moles × molar mass
Mass = 5.73 mol × 46.068 g/mol
Mass = 263.97 g
Mass of solvent:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = 932.7 g - 263.97 g
Mass of solvent = 668.73 g
In Kg = 668.73 /1000 = 0.6687 Kg
Molality:
Molality = number of moles of solute / mass of solvent in Kg
Molality = 5.73 mol / 0.6687 Kg
Molality = 8.57 m
Considering the definition of molality , you obtain that the molality of a 5.73 M ethanol (C₂H₅OH) solution whose density is 0.9327 g/mL is 8.57 [tex]\frac{moles}{kg}[/tex].
Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
[tex]Molality=\frac{number of moles of solute}{kilogramof solvent}[/tex]
You have a 5.73 M ethanol (C₂H₅OH) solution whose density is 0.9327 g/mL.
Molarity is the number of moles of solute that are dissolved in a certain volume.
In this case, taking into account that the volume considered is 1 L, the number of moles of solute is 5.73 moles.
On the other side, density is the ratio of the weight (mass) of a substance to the volume it occupies. So, being 1 mL= 0.001 L, 0.9327 g/mL means that you have 0.9327 grams per 1 mL or 932.7 g per 1 L.
So, being the mass of solution calculated as number of moles multiplied by the molar mass, and being the mass of the solution 932.7 grams in 1 L, the mass of water is:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = mass of solution - number of moles× molar mass
Mass of solvent = 932.7 g - 5.73 mol× 46.068 g/mol
Mass of solvent = 932.7 g - 263.97 g
Mass of solvent = 668.73 g= 0.66873 kg
Then, the molality can be calculated as:
[tex]Molality=\frac{5.73 moles}{0.66873 kg}[/tex]
Solving:
molality= 8.57 [tex]\frac{moles}{kg}[/tex]
Finally, the molality of a 5.73 M ethanol (C₂H₅OH) solution whose density is 0.9327 g/mL is 8.57 [tex]\frac{moles}{kg}[/tex].
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