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Sagot :
The volume in litres occupied by 1.59 Kg of ethyl alcohol given the data from the question is 2.03 L
How to determine the mole of 78.5 g of ethyl alcohol
- Mass of ethyl alcohol = 78.5 g
- Molar mass of ethyl alcohol = 46.07 g/mol
- Mole of ethyl alcohol =?
Mole = mass / molar mass
Mole of ethyl alcohol = 78.5 / 46.07
Mole of ethyl alcohol = 1.7 mole
How to determine the mole of 1.59 Kg of ethyl alcohol
- Mass of ethyl alcohol = 1.59 Kg = 1.59 × 1000 = 1590 g
- Molar mass of ethyl alcohol = 46.07 g/mol
- Mole of ethyl alcohol =?
Mole = mass / molar mass
Mole of ethyl alcohol = 1590 / 46.07
Mole of ethyl alcohol = 34.5 moles
How to determine the new volume
- Initial volume (V₁) = 100 mL = 100 / 1000 = 0.1 L
- Initial mole (n₁) = 1.7
- New mole (n₂) = 34.5 moles
- New Volume (V₂) =.?
V₁ / n₁ = V₂ / n₂
0.1 / 1.7 = V₂ / 34.5
Cross multiply
1.7 × V₂ = 0.1 × 34.5
Divide both side by 1.7
V₂ = (0.1 × 34.5) / 1.7
V₂ = 2.03 L
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